Chemistry High School

configurate the following element using sub level and group each elemnt v(z=23),Ni(z=28),Cu(z=29),zn(z=30),cr(z=24),Mn(z=25)​

Answers

Answer 1

Answer:

To configure the electron sublevels and groups for each element, we can use the following format:

Element symbol: [Electron configuration] Sublevel: Group

V (Z=23): [Ar] 3d3 4s2 Sublevel: 3d, 4s Group: 5, 4

Ni (Z=28): [Ar] 3d8 4s2 Sublevel: 3d, 4s Group: 10, 4

Cu (Z=29): [Ar] 3d10 4s1 Sublevel: 3d, 4s Group: 11, 4

Zn (Z=30): [Ar] 3d10 4s2 Sublevel: 3d, 4s Group: 12, 4

Cr (Z=24): [Ar] 3d5 4s1 Sublevel: 3d, 4s Group: 6, 4

Mn (Z=25): [Ar] 3d5 4s2 Sublevel: 3d, 4s Group: 7, 4

Related Questions

Calculate the pH at the following points in a titration of 40 mL (0. 040 L) of 0. 100 M

barbituric acid (Ka = 9. 8  10–5) with 0. 100 M KOH.

(a) no KOH added

(b) 20 mL of KOH solution added

(c) 39 mL of KOH solution added

(d) 40 mL of KOH solution added

(e) 41 mL of KOH solution added

Answers

We are given:

Molarity of
C
4
H
4
N
2
O
3
=
0.100

M
Volume of
C
4
H
4
N
2
O
3
=
40

m
L
Molarity of
K
O
H
=
0.100

M
K
a
=
9.8
×
10
−
5


Chemical reaction:
C
4
H
4
N
2
O
3
+
K
O
H
→
C
4
H
3
N
2
O
3
K
+
H
2
O



Part a:
no KOH is added.

The concentration of hydrogen ion can be calculated as:

C
4
H
4
N
2
O
3
⇌
C
4
H
3
N
2
O
−
3
+
H
+
K
a
=
[
C
4
H
3
N
2
O
−
3
]
[
H
+
]
[
C
4
H
4
N
2
O
3
]
[
C
4
H
3
N
2
O
−
3
]
=
[
H
+
]
9.8
×
10
−
5
=
[
H
+
]
2
[
C
4
H
4
N
2
O
3
]
9.8
×
10
−
5
=
[
H
+
]
2
0.100
[
H
+
]
2
=
9.8
×
10
−
6
[
H
+
]
=
3.13
×
10
−
3

M

The pH value of the solution is:

p
H
=
−
log
[
H
+
]
p
H
=
−
log
(
3.13
×
10
−
3
)
p
H
=
2.50



Part b:
20 mL of KOH is added.

We will find the initial moles of the reactants:

M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100

M
×
0.040

L
=
0.0040

m
o
l
n
K
O
H
=
0.100

M
×
0.020

L
=
0.0020

m
o
l

Moles of the acid and salt after the addition of KOH:

n
C
4
H
4
N
2
O
3
=
0.00400
−
0.0020
=
0.0020

m
o
l
n
C
4
H
3
N
2
O
3
K
=
0.00200

m
o
l

The concentration of acid and salt:

M
=
n
V
V
=
40
+
20
=
60

m
L
[
C
4
H
4
N
2
O
3
]
=
0.0020

m
o
l
0.060

L
=
0.033

M
[
C
4
H
3
N
2
O
3
K
]
=
0.0020

m
o
l
0.060

L
=
0.033

M

The pH value is calculated as:

p
H
=
−
log
(
K
a
)
+
log
(
[
C
4
H
3
N
2
O
3
K
]
[
C
4
H
4
N
2
O
3
]
)
p
H
=
−
log
(
9.8
×
10
−
5
)
+
log
(
0.033
0.033
)
p
H
=
4.01



Part c:
39 mL of KOH is added.

We will find the initial moles of the reactants:

M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100

M
×
0.040

L
=
0.0040

m
o
l
n
K
O
H
=
0.100

M
×
0.039

L
=
0.0039

m
o
l

Moles of the acid and salt after the addition of KOH:

n
C
4
H
4
N
2
O
3
=
0.00400
−
0.0039
=
0.0001

m
o
l
n
C
4
H
3
N
2
O
3
K
=
0.0039

m
o
l

The concentration of acid and salt:

M
=
n
V
V
=
40
+
39
=
79

m
L
[
C
4
H
4
N
2
O
3
]
=
0.0001

m
o
l
0.079

L
=
0.00126

M
[
C
4
H
3
N
2
O
3
K
]
=
0.0039

m
o
l
0.079

L
=
0.0494

M

The pH value is calculated as:

p
H
=
−
log
(
K
a
)
+
log
(
[
C
4
H
3
N
2
O
3
K
]
[
C
4
H
4
N
2
O
3
]
)
p
H
=
−
log
(
9.8
×
10
−
5
)
+
log
(
0.0494
0.00126
)
p
H
=
4.01
+
1.59
p
H
=
5.60



Part d:
40 mL of KOH is added.

We will find the initial moles of the reactants:

M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100

M
×
0.040

L
=
0.0040

m
o
l
n
K
O
H
=
0.100

M
×
0.040

L
=
0.0040

m
o
l

Moles of the acid and salt after the addition of KOH:

n
C
4
H
4
N
2
O
3
=
0.00400
−
0.0040
=
0.0

m
o
l
n
C
4
H
3
N
2
O
3
K
=
0.0040

m
o
l

The concentration of acid and salt:

M
=
n
V
V
=
40
+
40
=
80

m
L
[
C
4
H
4
N
2
O
3
]
=
0.0

m
o
l
0.080

L
=
0.0

M
[
C
4
H
3
N
2
O
3
K
]
=
0.0040

m
o
l
0.080

L
=
0.050

M

The pH value is calculated as:

p
H
=
7
+
1
2
(
p
K
a
+
log
[
C
4
H
3
N
2
O
3
K
]
)
p
H
=
7
+
1
2
(
4.01
+
log
(
0.05
)
)
p
H
=
7
+
1
2
(
4.01
−
1.30
)
p
H
=
7
+
1.35
p
H
=
8.35



Part e:
41 mL of KOH is added.

We will find the initial moles of the reactants:

M
o
l
e
s
(
n
)
=
M
o
l
a
r
i
t
y
(
M
)
×
V
o
l
u
m
e
(
V
)
n
C
4
H
4
N
2
O
3
=
0.100

M
×
0.040

L
=
0.0040

m
o
l
n
K
O
H
=
0.100

M
×
0.041

L
=
0.0041

m
o
l

Moles of the base remains left over after the addition of KOH:

n
K
O
H
=
0.0041
−
0.0040
=
0.0001

m
o
l

The concentration of base:

M
=
n
V
V
=
40
+
41
=
81

m
L
[
K
O
H
]
=
0.0001

m
o
l
0.081

L
=
0.00123

M

The concentration of hydroxide ion is equal to 0.00123 M because it dissolves as:

K
O
H
→
K
+
+
O
H
−

The pOH value of the solution:

p
O
H
=
−
log
[
O
H
−
]
p
O
H
=
−
log
(
0.00123
)
p
O
H
=
2.91

The pH value is calculated as:

p
H
+
p
O
H
=
14
p
H
+
2.91
=
14
p
H
=
11.1

14. What is the molecular mass of a substance if

22. 5 g dissolved in 250 g of water produces a

solution whose freezing point is -0. 930°C?

Answers

The molecular mass of the substance is 181 g/mol if 22. 5 g dissolved in 250 g of water produces a solution whose freezing point is -0. 930°C.

To determine the molecular mass of the substance, we can use the freezing point depression formula:

ΔTf = Kf·m

Where ΔTf is the freezing point depression, Kf is the freezing point depression constant of water (1.86°C·kg/mol), and m is the molality of the solution, which is the number of moles of solute per kilogram of solvent.

We can start by calculating the molality of the solution:

molality = moles of solute/mass of solvent (in kg)

Since we know that 22.5 g of the substance is dissolved in 250 g of water, we can calculate the mass of the solvent as:

mass of solvent = 250 g / 1000 = 0.250 kg

The mass of the solute can be calculated as:

mass of solute = 22.5 g / 1000 = 0.0225 kg

Now we can calculate the molality of the solution:

molality = 0.0225 kg / 0.250 kg = 0.09 mol/kg

Next, we can use the freezing point depression formula to calculate the molecular mass of the substance:

ΔTf = Kf·m

-0.930°C = 1.86°C·kg/mol x 0.09 mol/kg

Solving for the molecular mass (M):

M = (Kf x m) / ΔTf = (1.86°C·kg/mol x 0.09 mol/kg) / 0.930°C

M = 0.181 kg/mol = 181 g/mol

Therefore, the molecular mass of the substance is 181 g/mol.

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The particles in.......... can be separated from
heterogeneous mixtures by passing the mixture through a filter.
-suspension
-solution
-colloid
-pure substance

Answers

The particles in suspension can be separated from heterogenous mixtures by passing the mixture through a filter. Option 1.

Particles in a suspension

The particles in suspension can be separated from heterogeneous mixtures by passing the mixture through a filter.

A suspension is a heterogeneous mixture in which the particles are large enough to settle out over time and can be separated by physical means such as filtration.

Other options such as solution, colloid, and pure substances cannot be separated using a filter.

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What is the mass, in kilograms, of 3.28 x 103 mol of H2O?

Answers

3.28 x 10³ mol of water has a mass of 59.1156 kg. This is the accurate answer to the given question.

How do you determine 3.28 x 10³ mol of H2O's mass in kilograms?

We can use the molar mass of water, which is 18.02 g/mol, to calculate the mass of 3.28 x 10³ moles of H2O. The following equation can be used to translate the quantity of moles into grams:

mass is determined by multiplying the number of moles by the molar mass.

Inputting the values provided yields:

18.02 g/mol times 3.28 x 10 3 moles of mass equals 59,115.6 g.

Lastly, by dividing by 1000, we can change the mass from grams to kilograms:

mass is 59,115.6 grams per kilogram (59.1156 kg).

As a result, 3.28 x 10³ moles of H2O have a mass of 59.1156 kg.

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11. Which one of the following contains the same number of atoms as 7 g

of iron?

(a) 4 g of aluminium (b) 4g of magnesium (C) 4g of sulphur

(d) 3 g of carbon (e) 4 g of calcium

Answers

Answer:

4 grams of sulfur contains the same number of atoms as 7 g of iron

Explanation:

The statement "same number of atoms" is the same as stating "the same numer of moles."  Moles are a count of atoms/molecules/etc.

So let's convert the masses into moles.  Take the mass of an element and divide that by its molar mass (g/mole) to obtain moles of that element.

See the attached worksheet for these calculations.

The 7 grams of iron represents 0.125 moles of iron.  The worksheet tells us that 4 grams of sulfur is also 0.125 moles.  Nothing else has that number.  

7 g of iron contains the same number of atoms as 4 grams of sulfur.

 [1 mole = 6.02x10^23 particles].

What is the percent composition of MgSO4 in a solution if 18.6 grams MgSO4 is dissolved in 229 g water ?

Answers

Percent of water of crystallisation in MgSO4⋅7H2O=2467×18×100=51. 2 %.

How can you tell what proportion of the composition is water?

Divide by 100 the mass of the hydrate and the volume of water lost. The amount of water in one mole of the hydrate must be multiplied by 100 after being divided by the hydrate's molar mass in order to determine the theoretical (actual) percent hydration (percent water) using the formula.

Hence, this ratio might be represented by one magnesium sulfate formula unit to seven water molecules. Since MgSO47H2O is the chemical formula for magnesium sulfate hydrate, the value of x must be seven. The value of x in the chemical formula for magnesium sulfate hydrate is thus seven.

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is the diamond cubic structure close-packed? explain your rationale. note: a close-packed structure is one where you can find a plane of atoms in that structure where all the atoms touch. for example, fcc and hcp are close- packed structures.

Answers

Yes, the diamond cubic structure is close-packed. In a close-packed structure, you can find a plane of atoms where all the atoms touch. To understand why the diamond cubic structure is close-packed, follow these steps:

1. Identify the structure: The diamond cubic structure consists of two interpenetrating face-centered cubics (fcc) lattices with a relative shift of 1/4 along the space diagonal.

2. Locate the planes: In the diamond cubic structure, there are several planes where atoms touch each other, forming a hexagonal close-packed (hcp) arrangement. For instance, the (111) plane in one of the fcc sublattices is close-packed.

3. Examine neighboring atoms: In these close-packed planes, the atoms are arranged so that each atom touches its neighboring atoms. This close arrangement of atoms maximizes the packing efficiency.

So, based on the presence of close-packed planes and the neighboring atoms touching each other, the diamond cubic structure is considered a close-packed structure.

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-. A 100.0 g Chunk of Aluminum with an Initial Temperature of 450.0 C° is added to 100.0
mL of Ethyl Alcohol with an Initial Temperature of 80.0 C° (A) Calculate Equilibrium
Temperature of the Mixture (B) Calculate the Heat Exchange of the System
BA) 198 C° B) 22770 J All answers Approx
D A) 110C° B) 10000 J All answers Approx
A) 150 C° B) 30000 J All answers Approx
A) 300 C° B) 15000 J All answers Approx

Answers

The negative sign means that heat was transferred from the aluminium piece to the ethyl alcohol. Consequently, the heat exchange of the system is roughly -22,770 J, or -2.28 x 10^4 J (to 2 significant figures).

What happens when 100g of 100 C boiling water is introduced to a calorimeter?

The mixture's temperature rises to 20 C. The mixture in the calorimeter is then dipped into by a metallic block of mass 1 kilogramme at 10°C. The temperature rises to 19 C once thermal equilibrium has been reached.

Q = m * c * T, where Q is the heat exchanged, m is the mass of the substance, c is the specific heat capacity, and T is the change in temperature, is the formula used to determine the heat exchanged.

We can use the following formula to determine the equilibrium temperature:

m_al * c_al * (T_eq - T_al) = m_et * c_et * (T_et - T_eq)

By entering the specified values, we obtain:

(0.100 kg) * (0.902 J/g°C) * (T_eq - 450.0°C) = (0.100 kg) * (2.44 J/g°C) * (80.0°C - T_eq)

Simplifying, we get:

90.2 J/C * (T_eq - 450.0) = 244 J/C * (80.0 - T_eq)

90.2 T_eq - 40590 = 19520 - 244 T_eq

334.2 T_eq = 60110

T_eq ≈ 179.7°C ≈ 180°C (to the nearest 10°C)

As a result, the mixture's equilibrium temperature is roughly 180 °C.

(B) We can apply the same formula as before to determine the system's heat exchange. We can suppose that the mixture has a specific heat capacity equal to that of ethyl alcohol.

Q = m_al * c_al * (T_eq - T_al) + m_et * c_et * (T_eq - T_et)

Plugging in the given values, we get:

Q = (0.100 kg) * (0.902 J/g°C) * (180.0°C - 450.0°C) + (0.100 kg) * (2.44 J/g°C) * (180.0°C - 80.0°C)

Q ≈ -22,770 J ≈ -2.28 x 10⁴ J (to 2 significant figures)

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An excess of sodium carbonate, Na2CO3, in solution is added to a solution containing 19. 25 g CaCl2. After performing the experiment, 11. 13 g of calcium carbonate, CaCO3, is produced. Calculate the percent yield of this reaction

Answers

From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃
Next, we shall determine the theoretical yield of of CaCO₃. This can be obtained as follow:
From the balanced equation above,
111 g of CaCl₂ reacted to produce 100 g of CaCO₃.
Therefore,
15.71 g of CaCl₂ will react to produce = = 14.15 g of CaCO₃.
Thus, the theoretical yield of of CaCO₃ is 14.15 g
Finally, we shall determine the percentage yield. This can be obtained as follow:
Actual yield of CaCO₃ = 13.19 g
Theoretical yield of CaCO₃ = 14.15 g
Percentage yield =?

= 93.2%
Therefore, the percentage yield of the reaction is 93.2%

Two elements, and , combine to form two binary compounds. In the first compound, 23. 3 g of combines with 3. 00 g of. In the second compound, 7. 00 g of combines with 4. 50 g of. Show that these data are in accord with the law of multiple proportions. If the formula of the second compound is , what is the formula of the first compound?

Compound I:

Answers

The formula for the first Compound I is likely A2B, or O2N.

To show that the data are in accordance with the law of multiple proportions, we need to calculate the ratios of the masses of element B that combine with a fixed mass of element A in each compound. If these ratios are in simple whole-number ratios, then the law of multiple proportions is upheld.

For Compound I:

Mass of A (23.3 g) / Mass of B (3.00 g) = 7.77

For Compound II:

Mass of A (7.00 g) / Mass of B (4.50 g) = 1.56

These ratios are not equal, indicating that the two compounds have different ratios of A and B. However, we can simplify the ratio for Compound II by dividing both masses by the mass of B:

Mass of A (7.00 g / 4.50 g) = 1.56

This suggests that Compound II has a 2:3 ratio of A to B.

Using this information, we can determine the formula of Compound II. Let x be the atomic mass of A and y be the atomic mass of B:

(2x) + (3y) = 4.50 g

Solving for y gives us y = 1.00 g/mol. Since the formula of Compound II is AB2, we know that the atomic mass of A is 2 g/mol. Therefore, A is likely oxygen (atomic mass 16 g/mol).

For Compound I, we know that A is oxygen (atomic mass 16 g/mol). Using the same approach as above, we can determine that B has an atomic mass of 14 g/mol. Therefore, the formula for Compound I is likely A2B, or O2N.

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H2 hydrogen filled balloon took 28. 4 hours to go flat. How long would it have taken for helium balloon to go flat?

Answers

Without knowing more about the individual balloon and its surroundings, it is impossible to estimate the precise amount of time it would take for a helium-filled balloon to deflate.

Nonetheless, based on helium's general characteristics and its slower rate of diffusion than hydrogen, we would anticipate that a helium-filled balloon will outlast a hydrogen-filled balloon under comparable circumstances.

Generally, the helium gas diffuses slowly as compared to the hydrogen gas and that's why many ballon these days are filled with helium gas. Helium gas has the larger atomic size and lower solubility in many materials.

To estimate the time it would take for a helium-filled balloon to go flat, we can use the same basic formula that governs the diffusion of gases through materials.

This formula, known as Fick's law of diffusion, describes the rate at which a gas diffuses through a material as proportional to the concentration gradient of the gas across the material, as well as to a constant known as the diffusion coefficient.

In summary it is very difficult to exactly state the time in which  helium balloon to go flat without knowing the environment and other important factors.

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Zinc hydroxide, Zn(OH)2, is precipitated from the tailings ponds of zinc mining operations before the water is released into local rivers. This reduces the concentration of Zn2+ ions in the wastewater. 5.00 kg of sludge containing zinc hydroxide is titrated against HCl solution and 1.50 L of 2.0 M acid is required to reach the endpoint and neutralize the zinc hydroxide. What mass of zinc is contained in the 5.00 kg sludge?

Answers

Answer:

First, we need to find the number of moles of HCl used in the titration:

moles HCl = M x V = 2.0 mol/L x 1.50 L = 3.00 mol

Since zinc hydroxide and HCl react in a 1:2 molar ratio according to the balanced chemical equation:

Zn(OH)2 + 2 HCl → ZnCl2 + 2 H2O

we know that twice as many moles of HCl are required to react with each mole of Zn(OH)2. Therefore, the number of moles of zinc hydroxide in the 5.00 kg sludge can be calculated as follows:

moles Zn(OH)2 = 1/2 x moles HCl = 1/2 x 3.00 mol = 1.50 mol

Finally, we can use the molar mass of Zn(OH)2 to convert the number of moles to mass:

mass Zn(OH)2 = moles Zn(OH)2 x molar mass Zn(OH)2

= 1.50 mol x 99.39 g/mol

= 149.1 g

Therefore, the mass of zinc contained in the 5.00 kg sludge is 149.1 grams.

Explanation:

1- Give an example of gas in liquid solution.

2- Give an example of solid in soild solution.

3-Give an example of gas in gas solution.


Answers

Answer:

1- Oxygen in water

2- Brass, bronze and sterling silver

3- Air

How many grams of NH3
form when 22.3 L
of H2(g)
(measured at STP) reacts with N2
to form NH3
according to this reaction?

N2(g)+3H2(g)→2NH3(g)

Answers

Answer:

11.2823 grams of NH3 are produced
Explanation:

No. of moles for H2 = 22.3/22.4 = 0.9955 moles

By calculating number of moles produced of NH3, by using ratios:

N2 + 3H2 → 2NH3
1    :   3    :    2
? : 0.9955:  ?

part value = 0.9955/3 = 0.33183
No. of moles for NH3 = 2 * 0.33183 = 0.6637 moles

mass of produced NH3 (Molar mass = 17 g/mol) = 0.6637*17 = 11.2823 grams

(d) If 48.0 g of NaCl react with 19.0 g of H₂SO4, what mass of Na2SO4 will be produced?​

Answers

Answer:

the mass of Na2SO4 produced is 27.6 g.

Explanation:

determine the balanced chemical equation for the reaction between NaCl and H₂SO4, and then use stoichiometry to calculate the mass of Na2SO4 produced.

The balanced chemical equation for the reaction is:

2 NaCl + H₂SO4 → Na2SO4 + 2 HCl

This equation tells us that 2 moles of NaCl react with 1 mole of H₂SO4 to produce 1 mole of Na2SO4.

First, we need to determine the number of moles of NaCl and H₂SO4 we have:

moles of NaCl = mass / molar mass = 48.0 g / 58.44 g/mol = 0.821 mol

moles of H₂SO4 = mass / molar mass = 19.0 g / 98.08 g/mol = 0.194 mol

Since H₂SO4 is the limiting reactant (it is present in less amount than required for complete reaction), we need to use the number of moles of H₂SO4 to calculate the number of moles of Na2SO4 produced:

moles of Na2SO4 = moles of H₂SO4 / 1 x (1 mole Na2SO4 / 1 mole H₂SO4) = 0.194 mol x (1/1) = 0.194 mol

Finally, we can use the molar mass of Na2SO4 to calculate the mass of Na2SO4 produced:

mass of Na2SO4 = moles of Na2SO4 x molar mass = 0.194 mol x 142.04 g/mol = 27.6 g

Therefore, the mass of Na2SO4 produced is 27.6 g.

an impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess agno3 formed 2.044 g of agcl. what percentage of nacl is in the impure sample?

Answers

The percentage of NaCl in the impure sample is 76.25%.

The given weight of an impure sample of table salt is 0.8421 g, and the weight of AgCl formed is 2.044 g. The percentage of NaCl in the impure sample can be determined as follows:

Calculations: Gram molecular weight of NaCl = 23 + 35.5 = 58.5 g

Number of moles of AgCl = weight of AgCl / gram molecular weight of AgCl= 2.044 / 143.5 = 0.0142 mol

The equation of the reaction between NaCl and AgNO3 is given as follows:

NaCl + AgNO3 → NaNO3 + AgCl

It can be seen from the above equation that 1 mole of NaCl gives 1 mole of AgCl.So, the number of moles of NaCl present in the sample is 0.0142 mol.

Percentage of NaCl in the sample = number of moles of NaCl present in the sample / number of moles of the impure sample × molar mass of NaCl × 100= (0.0142 / (0.8421 / 58.5)) × 100= 76.25%

Therefore, the percentage of NaCl in the impure sample is 76.25%.

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Potassium nitrate (KNO;) decomposes on heating to give potassium nitrite (KNO.) and oxygen (02).
When 4.04 g of KNOs is heated, 3.40 g of KNOz is produced
Use the law of conservation of mass to work out the mass of 0, produced.?

Answers

2.02 g mass of oxygen are produced during the decomposition of 4.04 g of potassium nitrate.

Law of conservation of mass calculation.

According to the law of conservation of mass, the total mass of the reactants in a chemical reaction is equal to the total mass of the products. Therefore, we can use this law to determine the mass of oxygen produced in the decomposition of potassium nitrate.

Let's assume that x grams of oxygen (O2) are produced during the decomposition of 4.04 g of potassium nitrate (KNO3). The balanced chemical equation for the decomposition of KNO3 is:

2 KNO3 --> 2 KNO2 + O2

From the equation, we can see that 2 moles of KNO3 produce 1 mole of O2. We can use this information to set up a proportion to solve for x:

2 mol KNO3 / 1 mol O2 = 4.04 g KNO3 / x g O2

Solving for x, we get:

x = (1 mol O2 / 2 mol KNO3) x (4.04 g KNO3) = 2.02 g O2

Therefore, 2.02 g of oxygen are produced during the decomposition of 4.04 g of potassium nitrate using law of conservation of mass.

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What is one possible reason dolphin dissapered

Answers

Explanation:

Foremost among the threats is fishing bycatch—marine animals incidentally caught and killed in fishing operations.

This is a serious crisis, not only for these charismatic animals, but also for the health of our oceans, where cetaceans play an irreplaceable role as apex predators.

A sample of gas of mass 2.929g occupies a volume of 426mL at 0°C and 1.00atm pressure.what is molecular weight of the gas?

Answers

Answer:

154 g/mole

Explanation:

We are given the mass of the gas, but we also need the number of moles the 2.929g represents.  Since we are provided the conditions of the gas, we can the Ideal Gas law to find the number of moles of the mystery gas.  

Ideal Gas Law:  PV = nRT, where P, V, and T are the pressure, volume, and temperature (temperature must be in degrees Kelvin), n is the moles, and R is the gas constant.

Let's choose the gas constant that has the same units as we were given.  R = 0.0820575 [L⋅atm⋅/(K⋅mol)] comes the closest, but we'll still need to convert ml to liters(L) and °C to °K:

  426mL = 0.426L

  0°C = 273.25   [add 273.15 to the Centrigrade value]

Let's rearrnage the ideal gas law to solve for n, the number of moles:

                           n = (PV/RT)

Now enter the data:

                            n = (1atm)(0.426L)/[(0.0820575 L⋅atm⋅/(K⋅mol))*(273.15°K)]

n = (1atm)(0.426L)/[(0.0820575 L⋅atm⋅/(K⋅mol))*(273.15°K)]   [Units that cancle are highlighted]

n = (1)(0.426)/[(0.0820575 /(mol))*(273.15)]  [We are left only with moles (mol)

n = (0.426)/(0.0820575)/(273.15) [1/1/mol]  [Move the only unit out (1/1/mol)]

n = (0.426)/(0.0820575)/(273.15) [1/1/mol] = 0.0190 moles

Note that the unit moves to the top, i.e., : 1/1/mol = mole

We have the mass and the number of moles.  Divide the two to obtain molar mass:

      (2.929g)/(0.0190 moles) = 154 g/mole   This is also the molecular weight.

[I don't know what is a gas at 0°C and has that molecular weight]

To earn full credit for your answers, you must show the appropriate formula, the correct substitutions , and your answer including the correct units.

In 2013, there were 2.6 x 106 people living in Paris. If 7.34 x 104 babies were born what was the crude birth rate?

Answers

The birth rate is  28.2 births per year

What is the birth rate?

The number of births per 1,000 persons in a particular population over a given time period is commonly used to determine the birth rate.

The following is the formula that can be used to calculate the birth rate:

(Number of births / Population) x 1,000 is the birth rate.

The birth rate can be obtained from;

Birth rate = (Number of births / Population) x 1,000

We know that the birth rate can be obtained from;

7.34 x 10^4 / 2.6 x 10^6  * 1000

= 28.2 births per year

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Cite the conditions to make the bulb light up

Answers

In order to make a bulb light up, certain conditions need to be met. Firstly, there must be a closed circuit, meaning that there is a continuous flow of current through the circuit. This can be achieved by connecting the bulb to a power source, such as a battery or a generator, and completing the circuit with wires.

Secondly, the bulb must be connected properly to the circuit. This involves connecting the positive and negative terminals of the bulb to the corresponding terminals of the power source or the wires.

Finally, the bulb must have the correct voltage and wattage rating. Using a bulb with a higher or lower rating than the power source can cause the bulb to either not light up or burn out quickly.

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PLEASE HELP
It's for Modeling Electron Configurations

Answers

The valence electrons of oxygen are; 2s2 2p4

The valence electrons of nitrogen are 2s2 2p3

This is how we can be able to know the number of valence electrons.

What are the valence electrons of Nitrogen and oxygen?

Valence electrons are the electrons in the outermost shell of an atom. They are the electrons involved in chemical reactions and bonding with other atoms. In the case of nitrogen, its electron configuration is 1s^2 2s^2 2p^3, which means it has two electrons in the first shell, two electrons in the second shell, and three electrons in the outermost p orbital, giving it a total of 5 valence electrons.

Oxygen has an electron configuration of 1s^2 2s^2 2p^4, which means it has two electrons in the first shell, two electrons in the second shell, and four electrons in the outermost p orbital, giving it a total of 6 valence electrons.

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An aerosol deodorant can has a pressure of 3.00 atm at 25 °C. What is the pressure inside the can at a temperature of 845 °C? This example illustrates why you shouldn’t incinerate aerosol cans.

Answers

The pressure inside the aerosol can at 845°C is 33.8 atm. This is why it is extremely dangerous to incinerate aerosol cans, as they can explode when exposed to high temperatures.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the final pressure, volume, and temperature.

We are given the following:

Initial pressure P1 = 3.00 atm

Initial temperature T1 = 25°C = 298 K

We need to find the final pressure P2 at T2 = 845°C = 1118 K.

Since the can is sealed, volumes V1 and V2 remain the same.

Plugging these values into the combined gas law, we get:

(P1 x V1) / T1 = (P2 x V2) / T2

(3.00 atm x V1) / 298 K = (P2 x V1) / 1118 K

Simplifying this equation, we get the following:

P2 = (3.00 atm x V1 x 1118 K) / (298 K x V1)

P2 = 33.8 atm

Therefore, the pressure inside the aerosol can at 845°C is 33.8 atm, which is significantly higher than at room temperature. This demonstrates why it is extremely dangerous to incinerate aerosol cans, as they can explode when exposed to high temperatures.

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Predict whether each of the following reactions will occur in aqueous solutions. If you predict a reaction will occur, state the products formed and their states (s, l, or aq) and write a balanced chemical reaction. If you predict that a reaction will not occur, state your reasoning. Note: barium sulfate and silver bromide are insoluble salts.

a. Sodium hydroxide(aq) + ammonium sulfate(aq)
b. Niobium(V) sulfate(aq) + barium nitrate(aq)

Answers

a. Sodium hydroxide(aq) + ammonium sulfate(aq) will react to form solid ammonium hydroxide and sodium sulfate(aq). b. Niobium(V) sulfate(aq) + barium nitrate(aq) will react to form solid barium sulfate(s) and niobium(V) nitrate(aq).

What is the significance of predicting whether a reaction will occur in aqueous solution?

Predicting whether a reaction will occur in aqueous solution is important in understanding chemical reactions and their potential outcomes. It allows chemists to determine what products will form and whether a reaction is feasible or not based on the solubility of the reactants and products.

How do you predict whether a reaction will occur in aqueous solution?

To predict whether a reaction will occur in aqueous solution, one can use solubility rules to determine whether the products of the reaction will form an insoluble precipitate or remain in solution. If a precipitate forms, it indicates that a reaction has occurred, whereas if all the products remain in solution, the reaction will not occur. Additionally, one can use other chemical principles, such as redox reactions and acid-base reactions, to predict whether a reaction will occur.

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_________________________________ are scientists who study the processes that make cells work.

Answers

Answer:

Microbiologists

Explanation:

Microbiologists are scientists who study the processes that make cells work.

Two platinum plates are covered with an unknown metal. The mass of these new plates is the same. One of these plates is dipped into a mercury (II) sulphate solution, the other is dipped into a copper (II) sulphate solution. By the time all the metal on platinum had changed, the mass of the plate decreased by 3. 600% in the case of the copper (II) salt, and increased by 6. 675% in the case of the mercury (II) salt. Estimate the standard potential of the unknown metal and determine the metal in question.


Given: standard reduction potential of (Hg2+/Hg) and

(Cu2+/Cu)

Answers

By comparing the standard potentials for each reaction, we can estimate the standard potential of the unknown metal and determine the metal in question. To estimate the standard potential of the unknown metal, we need to use the Nernst equation:

[tex]E = Eo - (RT/nF) ln(Q)[/tex]


Where E is the standard potential, [tex]Eo[/tex] is the standard reduction potential, R is the gas constant, T is the temperature, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.
First, we need to calculate the change in mass for each of the platinum plates. For the plate dipped in the copper (II) sulphate solution, the change in mass is:
Δ[tex]m = (3.600/100) * m = 0.036 * m[/tex]
For the plate dipped in the mercury (II) sulphate solution, the change in mass is:
Δ[tex]m = (6.675/100) * m = 0.06675 * m[/tex]
Next, we need to calculate the reaction quotient (Q) for each reaction. The reaction quotient is the ratio of the concentrations of the products to the concentrations of the reactants. For the copper (II) sulphate solution, the reaction quotient is:
[tex]Q = [Cu2+]/[Cu][/tex]
For the mercury (II) sulphate solution, the reaction quotient is:
[tex]Q = [Hg2+]/[Hg][/tex]
Now, we can plug these values into the Nernst equation to estimate the standard potential of the unknown metal. For the copper (II) sulphate solution:
[tex]E = Eo - (RT/nF) ln(Q)\\= Eo - (RT/nF) ln([Cu2+]/[Cu]) \\= Eo - (RT/nF) ln(0.036)[/tex]
For the mercury (II) sulphate solution:
[tex]E = Eo - (RT/nF) ln(Q) \\= Eo - (RT/nF) ln([Hg2+]/[Hg]) \\= Eo - (RT/nF) ln(0.06675)[/tex]

We can estimate the standard potential of the unknown metal and identify it by comparing the standard potentials for each reaction.

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Is the pattern of atomic radius absolute or general (always true or generally true)?

Answers

Answer:  

Generally true.

Explanation:

As you go down the columns of the Periodic Table, the radius of the atom increases.

As you go across the Periodic Table (left to right), the radius of the atoms decrease.

Therefore, the second element on the table, Helium, has the smallest radius and element 87, Francium has the largest radius.

There are a few exception scattered throughout the table, such as the lanthanides and actinides, but generally speaking the radius increases as you travel from right to left and top to bottom.

The equation below represents a chemical reaction that occurs in living cells.
C6H12O6 + 6O2 ------> 6CO2 + 6H2O + energy

The reactants contain a total of _____?
atoms.

Answers

Answer:

36 atoms

Explanation:

The reactants in the equation contain 6 carbon atoms, 12 hydrogen atoms, and 18 oxygen atoms. So the total number of atoms in the reactants is:

6 (carbon atoms) + 12 (hydrogen atoms) + 18 (oxygen atoms) = 36 atoms

Copper react with oxygen to form two oxide x and y , on analysis 1. 535g of x yielded 1. 160g of copper so deter

mine the chemical formula of x and y

Answers

Copper reacts with oxygen to form two oxides, x and y; 1. 535g of x yielded 1. 160g of copper in analysis.the chemical formula of x and y is Cu2O3.

the amount of oxygen in oxide is

1.535 g of oxide x contains 1.160 g of copper.

Therefore, the mass of oxygen in oxide x is:

1.535 g - 1.160 g = 0.375 g

moles of oxygen = 0.375 g / 16.00 g/mol = 0.0234 mol

moles of copper = 1.160 g / 63.55 g/mol = 0.0182 mol

Now, we can divide both moles by the smaller one to get the mole ratio:

0.0234 mol / 0.0182 mol = 1.28

0.0182 mol / 0.0182 mol = 1.00

total mass of copper and oxygen in both oxides = 1.535 g + unknown mass of oxide y

Mass of oxygen in oxide y = mass of copper and oxygen in both oxides - mass of copper in oxide x

mass of oxygen in oxide y = (1.535 g + unknown mass of oxide y) - 1.160 g

mass of oxygen in oxide y = 0.375 g + unknown mass of oxide y

mass of oxide y = 1.535 g - 0.375 g = 1.160 g

oxide y has a mass of 1.160 g. Since the total mass of copper and oxygen in oxide y is the same as in oxide x, the formula for oxide y is Cu2O3.

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Explain in a three-paragraph essay the mechanics of how a battery works. How does the choice of metals used in a battery affect its performance? what specific metals work best?

Answers

A battery is a device that converts chemical energy into electrical energy through a process known as an electrochemical reaction.

How does a battery work ?

When a battery is connected to a circuit, the electrochemical reaction causes a flow of electrons from the anode to the cathode, generating an electric current that can power a device.

The metal chosen for the anode must be capable of losing electrons easily, while the metal chosen for the cathode must be capable of accepting electrons. The choice of metals can also affect the voltage and capacity of the battery, as well as its overall efficiency.

In general, the metals used in a battery should have a large difference in their electronegativity values, which determines how easily an atom can attract electrons. Common metals used in batteries include zinc, lithium, nickel, and cadmium.

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