Calculate the mass in 0.523 moles of Ag?

Answer:

3.48 x 1013 N2 molecules

step-by-step explanation:

Lets set up our equation first

P = 1.00 x 10-6 mm Hg T = 0.0° C + 273 = 273 K

We are given the V = 985 mL ,

R = 0.0821 L·atm/mol·K

Now use the ldeal gas law, but we are solving n, amount of substanvce

PV = nRT, we will change this equation to ;

n = PV/RT

n = 1.00 x 10-6 mm x 1 atm/760 mm x 985 mL x 1 L/103

mL/

(0.0821 L·atm/mol·K x 273 K) = 5.78 x 10-11 moles N2

nmolecules = 5.78 x 10-11 moles N2 x 6.02 x 1023 N2 molecules/1 mol N2

Answer:

It's a representative metal

Explanation: Transitional metals are metals of various chemical elements and have valence electrons—i.e., electrons that can participate in the formation of chemical bonds.

Answer:

Explanation:

Here you go!! Hope this helps.

Answer:

266Lr

Thirteen isotopes of lawrencium are currently known; the most stable is 266Lr with a half-life of 11 hours, but the shorter-lived 260Lr (half-life 2.7 minutes) is most commonly used in chemistry because it can be produced on a larger scale.

Explanation:

hopefully that helps you

Answer:

plucking is usually extracting hair

Explanation:

example: pluck the chicken feathers

pluck my brows

Answer:

The correct answer is 146 g/mol

Explanation:

Freezing point depression is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:

ΔTf = Kf x m

Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:

ΔTf = 1.02ºC

Kf = 5.12ºC/m

From this, we can calculate the molality:

m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m

The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:

0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute

There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:

molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol

Therefore, the molar mass of the compound is 146 g/mol

Answer:

3.80 × 10²⁴ atoms of F

Explanation:

Step 1: Given data

Moles of barium fluoride (BaF₂): 3.16 mol

Step 2: Calculate the moles of fluorine in 3.16 moles of barium fluoride

The molar ratio of BaF₂ to F is 1:2. The moles of F in 3.16 moles of BaF₂ are 2/1 × 3.16 mol = 6.32 mol.

Step 3: Calculate the atoms of F in 6.32 moles of F

We will use Avogadro's number: there are 6.02 × 10²³ atoms of F in 1 mole of atoms of F.

6.32 mol × (6.02 × 10²³ atom/1 mol) = 3.80 × 10²⁴ atom

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of [tex]Na[/tex] = 5.0 g

Mass of [tex]Cl_2[/tex] = 10.0 g

Molar mass of [tex]Na[/tex] = 23 g/mol

Molar mass of [tex]Cl_2[/tex] = 71 g/mol

First we have to calculate the moles of [tex]Na[/tex] and [tex]Cl_2[/tex].

[tex]\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}[/tex]

[tex]\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol[/tex]

and,

[tex]\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}[/tex]

[tex]\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

[tex]2Na+Cl_2\rightarrow 2NaCl[/tex]

From the balanced reaction we conclude that

As, 2 mole of [tex]Na[/tex] react with 1 mole of [tex]Cl_2[/tex]

So, 0.217 moles of [tex]Na[/tex] react with [tex]\frac{0.217}{2}=0.108[/tex] moles of [tex]Cl_2[/tex]

From this we conclude that, [tex]Cl_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Na[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]NaCl[/tex]

From the reaction, we conclude that

As, 2 mole of [tex]Na[/tex] react to give 2 mole of [tex]NaCl[/tex]

So, 0.217 mole of [tex]HCl[/tex] react to give 0.217 mole of [tex]NaCl[/tex]

Now we have to calculate the mass of [tex]NaCl[/tex]

[tex]\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl[/tex]

Molar mass of [tex]NaCl[/tex] = 58.5 g/mole

[tex]\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g[/tex]

Now we have to calculate the percent yield of this reaction.

Percent yield = [tex]\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100[/tex]

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = [tex]\frac{5.24g}{12.7g}\times 100[/tex]

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

Answer:

Moles of  H₂ left = 0.051 mol

Moles of ammonia formed = 0.34 mol

Explanation:

Given data:

Moles of N₂ = 0.170 mol

Moles of H₂ = 0.561 mol

Moles of ammonia formed = ?

Moles of H₂ left = ?

Solution:

Balance chemical equation:

N₂ + 3H₂       →     2NH₃

Now we will compare the moles of ammonia with nitrogen because nitrogen is limiting reactant and limit the yield of ammonia.

                  N₂          :           NH₃

                     1           :              2

                   0.170     :            2×0.170 = 0.34 mol

Moles of ammonia formed = 0.34 mol

Moles of H₂ reacted:

                   N₂           :            H₂

                    1             :             3

                  0.170       :            3/1×0.170 = 0.51 mol

0.51 moles of hydrogen react with 0.170 moles of nitrogen.

Moles of H₂ left:

Moles of H₂ left = Total - moles reacted

Moles of  H₂ left = 0.561 mol - 0.51 mol

Moles of  H₂ left = 0.051 mol

Answer:

An element is atoms with the same number of protons.

Explanation:

Protons, electrons, and neutrons.

Answer:

35.1% is percent yield

Explanation:

Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

Moles Ni (Molar mass: 58.69g/mol):

1.02g * (1mol / 58.69g) = 0.01738moles Ni

Moles HBr:

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

Where P is pressure (0.0910atm)

V is volume (2.50L)

R is gas constant (0.082atmL/molK)

T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)

And n are moles

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

Answer:

a.

Explanation:

One double bond.

For example ethene CH2=CH2

Answer:

In this example, a 3 kilogram mass, at a height of 5 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s 2 * 5m = 147.15 J. 9.81 meters per second squared (or more accurately 9.80665 m/s 2 ) is widely accepted among scientists as a working average value for Earth's gravitational pull.

Explanation:

Answer:

[tex]k=6.48x10^{-3}M^{-1}s^{-1}[/tex]

Explanation:

Hello.

In this case, based on the given information, we can write the rate law as shown below:

[tex]r=k[H_2O_2][KI][/tex]

Since the overall order of reaction is 2 being 1 for reach reactant. Thus, by knowing the initial rate and concentrations, the rate constant turns out:

[tex]k=\frac{r}{[H_2O_2][KI]}=\frac{7.50x10^{-4}M/s}{0.546M*0.212M}\\ \\k=6.48x10^{-3}M^{-1}s^{-1}[/tex]

Best regards!

The rate constant, k is [tex]6.48x10^{-3}M^{-1}s^{-1}[/tex]

The rate law should be

[tex]r = k[H_2O_2]{KI}[/tex]

Since the total order of the reaction is 2 being 1 for reach reactant. So, by knowing the beginning rate and concentrations, the rate constant turns out:

[tex]k = \frac{r}{k[H_2O_2]{KI}} = \frac{7.50x10^{-4}M/s}{0.546M\times 0.212M}[/tex]

k = [tex]6.48x10^{-3}M^{-1}s^{-1}[/tex]

Learn more about reactants here; https://brainly.com/question/21029530

Answer:

48000

Explanation:

maths

Answer:

48000

Explanation:

(5x10^3) = 5 × 1000 = 5000

(4.3x10^4) = 4.3 × 10000 = 43000

5000 + 43000 = 48000

Answer:

The correct answer is b.

Explanation:

hope this helps u

Answer:

Bryna is correct it is B

Explanation:

Answer:

Boron and Aluminum

Explanation:

If you write the electron configuration for boron and aluminum, you get:

[tex]1s^22s^22p^1[/tex] for boron and [tex]1s^22s^22p^63s^23p^1[/tex] for aluminum. Both have 3 valance electrons and has 2 electrons in a s-orbital and 1 in a p-orbital. These valance electron similarities are based on the column/group the elements are. Therefore, Boron and Aluminum have similar chemical behaviours and similar arrangement of outer/valance electrons.

Answer:

In carbon dioxide, on the right, there is 2.666 g of oxygen for every gram of carbon. So the ratio of oxygen in the two compounds is 1:2, a small whole number ratio

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Answer:

geosphere

Explanation:

geo means rocks this is the area

hope this helps dude

Answer:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + 3O₂(g)

Explanation:

Chemical equation:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + O₂(g)

Balance chemical equation:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + 3O₂(g)

Step 1:

Ca(ClO₃)₂(s)      →      CaCl₂(s) + O₂(g)

Left hand side                     Right hand side

Ca = 1                                     Ca = 1

Cl = 2                                     Cl = 2

O =  6                                     O = 2

Step 2:

 Ca(ClO₃)₂(s)      →      CaCl₂(s) + 3O₂(g)

Left hand side                     Right hand side

Ca = 1                                     Ca = 1

Cl = 2                                     Cl = 2

O =  6                                     O = 6

Answer:

0.296j/g⁰c

Explanation:

we have the following information from this question before us.

mass iv substance = 50grams

we have initial temperature ti = 20.2⁰c

final temperature = 125⁰c

the energy that was provided = 155kj

we proceed with this formula

energy = mcΔT

1.55x10³ = 50 x c x (125-20.2)

1.55x10³ = c x 50gm x 104.8k

we divide through to get c

c = 1.55x10³/50g x 104.8

c = 0.296J/g⁰c

that is the specific heat of this substance.

thank you!

Answer:

Explanation:

Kb of  (CH₃)₃N is 7.4 x 10⁻⁵

initial concentration of (CH₃)₃N   a   is .48 M

(CH₃)₃N    +   H₂O =  (CH₃)₃NH⁺  +  OH⁻

a - x                                     x               x  

x² / (a - x )  = Kb

x is far less than a so a - x can be replaced by a .

x² / a   = Kb

x²  = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶

x = 5.96 x 10⁻³

pOH = - log ( 5.96 x 10⁻³ )

= 3 - log 5.96

= 3 - .775

= 2.225

Answer:

The correct option is C

Explanation:

Chemical equilibrium is a state in which there is no net change in the amount of reactant and products formed over time. This reaction is a reversible reaction, hence the reaction can keep proceeding in either direction until the products formed are equal in mass to that of the reactants or the products been reversed back to the reactants until both sides (product and reactant) achieve equal mass.

Answer:

This question is incomplete

Explanation:

This question is incomplete but the answer to (1) and steps involved in calculating (2) and (3) will be discussed in the attachment

1. The reaction is an exothermic reaction because heat was given off. Exothermic reaction is a reaction in which heat is released from a reaction into the surroundings. And according to the question, the student observed that the temperature of the water increased during the course of the minutes.

Answer:

The concentration of chloride ion is [tex]2.82\times10^{-3}\;mol/L[/tex]

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the [tex]Cl^-[/tex] content 100 ppm suggests the presence of 100 mg of [tex]Cl^-[/tex] in 1 L of solution.

The molar mass of [tex]Cl^-[/tex] is equal to the molar mass of Cl atom as the mass of the excess electron in [tex]Cl^-[/tex] is negligible as compared to the mass of Cl atom.

So, the molar mass of [tex]Cl^-[/tex] is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of [tex]Cl^-[/tex] present in 100 mg (0.100 g) of [tex]Cl^-[/tex] is calculated as shown below:

[tex]N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol[/tex]

So, there is [tex]2.82\times10^{-3}\;mol[/tex] of [tex]Cl^-[/tex] present in 1 L of solution.