Consider the hypothetical thermochemical equation 3 A + B → 2 C for which ΔH = 65.5 kJ/mol.What would ∆H, in kJ/mol, be for the reaction 2 C → 3 A + B?

Answers

Answer 1
Answer:[tex]-65.5\text{kJ/mol}[/tex]

Explanations:

Given the hypothetical thermochemical equation, expressed as:

[tex]3A+B\rightarrow2C;\triangle H=65.5\text{kJ/mol}[/tex]

We are to find the enthalpy change (ΔH ) for the reaction 2C → 3 A + B.

• You can see that the, given reaction, 2C → 3 A + B is the, reverse ,of the thermochemical equation 3 A + B → 2 C.

• Since the ,reverse reaction is possible,, the reaction enthalpy has the ,same numerical value, but with the, ,opposite sign

Therefore, the enthalpy change ∆H, in kJ/mol, be for the reaction 2 C → 3 A + B is -65.5kJ/mol


Related Questions

What is the mass of 9.50 moles of magnesium chloride, MgCl2 ?'

Answers

The actual mass is 904.4g but with correct number of sig figs it’s 904g.

When 29.0 g of butane reacts with oxygen, 88.0 g of carbon dioxide and 45.0 g of water are produced. What mass of oxygen reacted with the butane?

Answers

We need to first write the balanced equation:

[tex]2C_4H_{10}+13O_2\rightarrow8CO_2+10H_2O[/tex]

we are given the following:

mass of butane = 29.0 g

mass of carbon = 88.0 g

mass of water = 45.0 g

We want the mass of O2 that reacted.

C4H10 is the limiting reactant, and we know the masses of products produced. We can use that to find out how much oxygen reacted.

We can use CO2:

number of moles of CO2 = 88.0/44.01 = 1.9995 mol

The molar ratio between O2 and CO2 is 13:8

Therefore the number of moles of O2 = 1.9995 x (13/8) = 3.249 mol

Now that we have the number of moles, we can calculate the mass.

m = n x M

m = 3.249 mol x 31.998 g/mol

m = 103.97 g

Definition: This is a type of element or substance that is not a metal.Example: oxygen, nitrogen, hydrogen

Answers

This elements or substances that are not a metal are called non-metals.

I think that the answer is Non-metal

What is the maximum number of electrons that can occupy the n = 2 shell?

Answers

Explanation:

If n = 2 we have this posibility:

This shell contains 2 subshells: s and p

Subshell 2s can only contain 2 electrons

Subshell 2p can only contain 6 electrons

Answer: In total, we can have 8 electrons in n = 2

Why is CuSO4 the correct formula for copper (ii) sulfate and not CuSO3?

Answers

The correct formula of copper (II) sulfate is CuSO₄ because Copper ion (Cu²⁺) has +2 charge and sulfate ion SO₄²⁻ has -2 charge, while CuSO₃ is copper(II) sulfite.

What is copper sulfate?

The CuSO₄ molecule contains an ionic bond between the copper cation (Cu²⁺) and the sulfate anion (SO₄²⁻).

The copper sulfate in its pentahydrate form is given by the chemical formula CuSO₄.5H₂O. This form of copper sulfate is characterized by its bright blue color but the anhydrous form of this salt is a white powder.

The oxidation state of the copper atom in a CuSO₄ molecule is +2. The molar mass of the pentahydrate and anhydrous forms of copper sulfate are 249.685 grams/mole and  159.609 grams/mole respectively.

Anhydrous CuSO₄ has a powdery appearance and a grey-white, while the pentahydrate has a bright blue color.

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How to balance ____CaCl2 —-> ____Ca+ ____ Cl

Answers

In order to properly balance an equation, we need to make sure that the same amount of elements on the reactants side matches the number of elements on the products side, we can do that by increasing the number in front of each molecule, the so called stoichiometric coefficient. In the reaction from the question we can properly balance by adding the following stoichiometric coefficients

For this question we have:

CaCl2 -> Ca + 2 Cl

What is the correct name-formula pair for calcium (II) oxide?

Answers

The chemical formula of calcium (II) oxide is CaO, wherein the valency of both oxygen and calcium is two.

What is chemical formula?

Chemical  formula is a way of representing the number of atoms present in a compound or molecule.It is written with the help of symbols  of elements. It also makes use of brackets and subscripts.

Subscripts are used to denote number of atoms of each element and brackets indicate presence of group of atoms. Chemical formula does not contain words. Chemical formula in the simplest form  is called empirical formula.

It is not the same as structural formula and does not have any information regarding structure.It does not provide any information regarding structure of molecule as obtained in structural formula.

There are four types of chemical formula:

1)empirical formula

2) structural formula

3)condensed formula

4)molecular formula

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How many grams of MgO are produced when 40.0 grams of O2 react completely with Mg? I picked B but I’m not sure if I’m correct

Answers

Step 1

The reaction provided:

2 Mg (s) + O2 (g) => 2 MgO (s) (completed and balanced)

-----------

Step 2

Information provided:

40.0 g of O2 which react completely

---

Information needed:

The molar masses of:

O2) 32.0 g/mol

MgO) 40.3 g/mol

-----------

Step 3

By stoichiometry,

2 Mg (s) + O2 (g) => 2 MgO (s)

32.0 g O2 ------- 2 x 40.3 g MgO

40.0 g O2 ------- X

X = 40.0 g O2 x 2 x 40.3 g MgO/32.0 g O2

X = 101 g MgO

Answer: C 101 g

Given the equation below, how many moles of nitrogen gas (N2) areneeded to react with 7.5 moles of hydrogen gas (H2)?N2+ 3H2 —>2 NH3

Answers

Answer

2.5 moles of N₂ are needed to react with 7.5 moles of hydrogen gas

Explanation

Given:

Equation: N₂ + 3H₂ → 2NH₃

Moles of H₂ = 7.5 moles

What to find:

The moles of nitrogen gas (N₂) needed to react with 7.5 moles of hydrogen gas

Step-by-step solution:

Let the mole of N₂ needed be x.

From the given balanced chemical equation:

3 moles of H₂ react with 1 mole of N₂

Therefore, 7. 5 moles of H₂ will react with x moles of N₂

Cross multiply

[tex]\begin{gathered} x\text{ moles }N_2\times3\text{ moles H}_2=7.5\text{ moles H}_2\times1\text{ mole N}_2 \\ \text{Divide both sides by 3 moles H}_2 \\ \frac{x\text{ moles }N_2\times3\text{ moles H}_2}{3\text{ moles H}_2}=\frac{7.5\text{ moles H}_2\times1\text{ mole N}_{2}}{3\text{ moles H}_2} \\ x\text{ moles }N_2=2.5\text{ moles} \end{gathered}[/tex]


1. 60.0 mL of 0.322 M lithium chloride, LICI (aq) are combined with 20.0 mL of 0.530 M Tin (II) nitrate,
Sn(NO₂)2 (aq), 0.632 g of precipitate are recovered.
a. Write a balanced chemical equation for the reaction.
b. Write a balanced net-ionic equation for the reaction.
c. Calculate the moles of precipitate that are actually produced in the reaction.
d. Calculate the moles of precipitate that should be produced if the reaction went to completion.
e. Calculate the percent yield of the reaction.

Answers

Answer:

60.0 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 M lead (II) nitrate.

How many grams of lead (II) iodide will precipitate? (you must write your own reaction)

Explanation:

After conditions changed to a volume of a sample of helium at 15.56 mL, 138.7°C and 334.6 kPa. What was its initial volyme at 63.2 °C and 57.3 kPa?O a. 74.2O b. 41.4O c. 2.18O d. 111

Answers

Answer

a. 74.2

Explanation

Given that:

The initial temperature, T₁ = 63.2 °C + 273 = 336.2 K

Initial pressure, P₁ = 57.3 kPa

The final volume, V₂ = 15.56 mL

Final temperature, T₂ = 138.7°C + 273 = 411.7 K

Final pressure, P₂ = 334.6 kPa

What to find:

The initial volume, V₁.

Step-by-step solution:

The initial volume, V₁ can be calculated using the combined gas law equation.

[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \\ V_1=\frac{P_2V_2T_1}{P_1T_2}=\frac{334.6kPa\times15.56mL\times336.2K}{57.3kPa\times411.7K} \\ \\ V_1=\frac{1750383.611\text{ }mL}{23590.41}=74.2\text{ }mL \end{gathered}[/tex]

Hence, its initial volume at 63.2 °C and 57.3 kPa is 74.2 mL

A student wants to produce a 1.8 M solution and has 0.9 moles of solute available. What is the maximum volume (in mL) of solution that can be produced?Given:Find:Equation used:Answer:

Answers

Answer:

[tex]500\text{ mL}[/tex]

Explanation:

Here, we want to get the maximum volume of solution that can be produced

Given:

Molarity = 1.8 M

Number of Moles = 0.9 moles

Find:

Volume

Equation Used:

Number of moles = molarity * volume

Answer:

[tex]\begin{gathered} 0.9\text{ = 1.8 }\times\text{ V} \\ V\text{ = }\frac{0.9}{1.8} \\ V=0.5dm^3 \end{gathered}[/tex]

To convert this to mL, we multiply the volume by 1000 since 1 L = 1000 mL

Thus, we have it that:

[tex]0.5\text{ }\times1000\text{ = 500 mL}[/tex]

A photon has a frequency of 3.16 * 10^7 Hz. Calculate its energy

Answers

If the photon has a frequency of   3.16 × 10 ⁷ Hz, then the energy of the the photon as per Planck's law would be 1.978× 10⁻²⁶ Joules

What is Wavelength?

It can be understood in terms of the distance between any two similar successive points across any wave.

C = λν

As given in the problem we have to calculate the energy of the light if the frequency of light is  3.16 × 10 ⁷ Hz

The energy of the photon = h ν

                                           = 6.26 × 10⁻³⁴ × 3.16 × 10 ⁷

                                           = 1.978× 10⁻²⁶ Joules

Thus, If the photon has a frequency of   3.16 × 10 ⁷ Hz, then the energy of the the photon as per Planck's law would be 1.978× 10⁻²⁶ Joules.

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Chemistry Electromagnetic Radiation HW Help. Pls real answers

Answers

The frequency of the green light from a traffic signal is 5.76 * 10¹⁴ Hz

The frequency of the light is 1.85 * 10¹⁶ Hz

The frequency of the light is 4.84 * 10¹⁴ Hz

The energy of the photons is 1.172 * 10⁻³ J

The energy of the photons is 4.34 * 10⁻² J

What is the frequency of light wave given the wavelength?

The frequency of light can be found given the wavelength from the formula below:

frequency = velocity / wavelength

8. The frequency of the green light from a traffic signal is calculated below:

velocity of light = 3.0 * 10⁸ m/s

wavelength = 5.2 x 10⁻⁷ m

Frequency = 3.0 * 10⁸ / 5.2 x 10⁻⁷

frequency = 5.76 * 10¹⁴ Hz

9. The frequency of the light is calculated below:

velocity of light = 3.0 * 10⁸ m/s

wavelength = 16.23 x 10⁻⁹ m

Frequency = 3.0 * 10⁸ / 16.23 x 10⁻⁹

frequency = 1.85 * 10¹⁶ Hz

10. The frequency of the light is calculated below:

velocity of light = 3.0 * 10⁸ m/s

wavelength = 6.2 x 10⁻⁷ m

Frequency = 3.0 * 10⁸ / 6.2 x 10⁻⁷

frequency = 4.84 * 10¹⁴ Hz

11. The energy of one of the photons will be calculated as follows:

E = hf

E = hc/λ

Energy of the photons =  2.3 x 10¹¹ x hc/λ

h = 6.626 * 10⁻³⁴Js

λ = 3.9 x 10⁻¹² m

c = 3.0 * 10⁸ m/s

E =   2.3 x 10¹¹ x 6.626 * 10⁻³⁴ * 3.0 * 10⁸ / 3.9 x 10⁻¹²

E = 1.172 * 10⁻³ J

12. The energy of one of the photons will be calculated as follows:

E = hf

E = hc/λ

Energy of the photons =  9.4 x 10¹⁶ x hc/λ

h = 6.626 * 10⁻³⁴Js

λ = 4.3 x 10⁻⁷ m

c = 3.0 * 10⁸ m/s

E =   9.4 x 10¹⁶ x 6.626 * 10⁻³⁴ * 3.0 * 10⁸ / 4.3 x 10⁻⁷

E = 4.34 * 10⁻² J

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In the reaction, 2NaOH + H2SO4 —> Na2SO4 + H2O, 40.0 g NaOH reacts with 60.0 g H2SO4. Which is the limiting reactant

Answers

Step 1

The reaction must be written, completed, and balanced:

2 NaOH + H2SO4 => Na2SO4 + H2O

-----------

Step 2

Information provided:

Mass of NaOH = 40.0 g

Mass of H2SO4 = 60.0 g

--

Information needed:

The molar masses:

NaOH) 40.0 g/mol approx.

H2SO4) 98.0 g/mol approx.

-----------

Step 3

The limiting reactant?

By stoichiometry:

1 mole NaOH = 40.0 g

1 mole H2SO4 = 98.0 g

2 NaOH + H2SO4 => Na2SO4 + H2O

2 x 40.0 g NaOH ----------- 98.0 g H2SO4

40.0 g NaOH ----------- X

X = 40.0 g NaOH x 98.0 g H2SO4/2 x 40.0 g NaOH = 49.0 g H2SO4

For 40.0 g of NaOH, 49.0 g of H2SO4 is needed but is provided 60.0 g of H2SO4. Therefore, the excess is the H2SO4, and the limiting reactant is the NaOH.

Answer: the limiting reactant is NaOH

Which temperature is unattainable?
a)_1k
b)_1°C
c)_1°F
d)All of these

Answers

Answer:

A is the correct answer.

Explanation:

It is possible to reach 1°C and 1°F because they are both units of temperature. But 1k is not a unit of temperature.

**NEED USEFUL ANSWER ASAP**
How do the masses of the hottest main sequence stars compare to the masses of the coolest main sequence stars?

Answers

Answer:

the more hotter the star, the more brightly it burns

Explanation:

someone just deleted my answer ugh sorry....ill write it again

so, te main sequence is a sequence in mass (and not a sequence in time).

The most massive stars are located at the the top left (since they are the brightest and hottest/bluest). The lowest mass stars are are located at the bottom left ( since they dimmer and cooler/redder).

Following the main sequence from the top left to the bottom right is thus a sequence from high to low mass.

Part II Dilution Problems: 17. 350.0 mL of water was added to a 2.3 L solution of NaCl. If the final concentration of the solution was 0.967 M, what was the original concentration of the solution?

Answers

ANSWER

The original concentration of the solution is 0.147 mol

EXPLANATION

Given that;

The volume of water is 350.0mL

The volume of NaCl solution is 2.3L

The final concentration of the solution is 0.967M

To find the original concentration of the solution, follow the steps below

Step1: Write the dilution formula

[tex]\text{ M1V1}=M2V2[/tex]

Where

• M1 is the original concentration of the solution

,

• V1 is the original volume of the of the solution

,

• M2 is the final concentration of the solution

,

• V2 is the final volume of the solution

Step 2: Convert the volume of water to L

Recall, that 1mL is equivalent to 0.001L

[tex]\begin{gathered} \text{ The volume can be converted below as} \\ \text{ 1mL }\rightarrow\text{ 0.001L} \\ \text{ 350mL }\rightarrow\text{ vL} \\ \text{ Cross multiply} \\ \text{ vL}\times\text{ 1mL }=\text{ 350mL}\times0.001L \\ \text{ Isolate v} \\ \text{ v }=\text{ }\frac{350\cancel{mL}\times\text{ 0.001L}}{1\cancel{mL}} \\ \text{ v }=\text{ 350}\times\text{ 0.001} \\ \text{ v }=\text{ 0.35L} \end{gathered}[/tex]

Step 3; Find the original concentration of the solution by substituting by the given data into the formula in step 1

[tex]\begin{gathered} \text{ M1}\times2.3\text{ }=\text{ 0.35}\times\text{ 0.967} \\ \text{ 2.3M1 }=\text{ 0.33845} \\ \text{ Divide both sides by 2.3} \\ \text{ }\frac{2.3M1}{2.3}\text{ }=\text{ }\frac{0.33845}{2.3} \\ \text{ M1}=\text{ 0.147 mol} \end{gathered}[/tex]

Hence, the original concentration of the solution is 0.147 mol

A lawyer is presenting a hair sample with its corresponding nuclear DNA that was found at the crime scene. How will this MOST likely affect the criminal case?

A.
The evidence will have a low probability of providing accurate matching results.

B.
The evidence will need to be looked at under a microscope to be admitted.

C.
The evidence will not be admissible in a court case.

D.
The evidence will provide a nearly exact match to the suspect’s hair.

Answers

A sample will likely affect a criminal case because this evidence will provide a nearly exact match to the suspect’s hair (Option D).

What is DNA testing?

DNA testing refers to the techniques used to obtain DNA from samples that can be used to identify an individual in a police case. A sample of hair may be used for DNA testing because it contains cells and therefore also contains DNA.

Therefore, with this data, we can see that DNA testing allow us to identify individuals for which we need to obtain a sample with cells.

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Monica works in a crime lab. Her supervisor tells her that she needs to better understand Locard’s Exchange Principle. How will she MOST accurately demonstrate her understanding of this concept?

A.
She will always test for the presence of blood at crime scenes.

B.
She will use both fingerprints and bullet matching to solve crimes.

C.
She will keep evidence separated so it doesn’t get contaminated.

D.
She will exchange information openly with other crime investigators.

Answers

She will keep evidence separated so it doesn’t get contaminated demonstrates the understanding of Locard’s Exchange Principle and is denoted as option C.

What is Locard’s Exchange Principle?

This principle states that an individual who commits a crime will bring something into the crime scene and leave with something from it. This also explains that they can be used as evidence during investigation by the appropriate authorities.

Since we are aware that new substances are always brought into and carried away from the crime scene then it is best to keep the evidence separated so it doesn’t get contaminated.

This ensures that the results gotten from the analysis and investigation are accurate and is therefore the reason why option C was chosen as the correct choice.

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Complete the table below by deciding whether a precipitate forms when aqueous solutions A and B are mixed. If a precipitate will form, enter its empirical formula in the last column.

Answers

1) A precipitate called [tex]Fe(OH)_{2}[/tex]

2) A precipitate is formed called [tex]Mg(CH_{3} COO)_{2}[/tex]

What is a precipitate?

The term precipitate is used to describe the product that is formed when there is a reaction between two aqueous phase reactants that leads to the formation of a solid product from the reaction as we can see from the image that is attached.

we now have to look at the reactions as we can see them in the mage that is attached. We must note that we can only say that a precipitate has been formed if the product is solid after we have mixed the aqueous phase reactants.

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Write formulas or names as appropriate for each of the following hydrates. 1. Magnesium sulfate heptahydrate 6. CoSO4•H2O 2. Copper(I) sulfate pentahydrate 7. Na2CrO4•4H2O 3. Potassium phosphate decahydrate 8. CuF2•2H2O 4. Calcium chloride hexahydrate 9. Sr(NO3)2•6H2O 5. Iron(III) nitrate nonahydrate 10. ZnSO4•7H2O

Answers

Answer:

Explanation:

Here, we want to give the names or formulas of the given hydrates

1) Magnesium sulfate heptahydrate

The first thing we have to do here is write the formula for magnesium sulfate.

Hepta means 7, which is pointing to the fact that there are 7 water molecules

We can have this as:

[tex]\text{MgSO}_4\cdot7H_2O[/tex]

6) Here, we can see that the element involved is cobalt with an oxideation number of 2. We also have just one water of crystallization

So, we have the name of the compound as:

[tex]\text{Cobalt (II) Sulphate hydrate}[/tex]

Balance the equation: __MgCl2 + __Cr2O3 --> __MgO + __CrCl3

Answers

We start by counting the number of atoms of each element on each side of the reaction.

Now we balance the reaction by trial and error, starting with chlorine. We have 2 chlorine atoms in the reactants and 3 clear atoms in the products, so to balance we cross the coefficients placing 3 on the reactant side and two on the product side in the respective molecules:

[tex]3Mg_{}Cl_2+Cr_2O_3\rightarrow MgO+2CrCl_3[/tex]

Now we continue with the oxygens, we have 3 oxygens on the reactant side and 1 on the products, so we put the coefficient 3 on the products side in front of the respective molecule:

[tex]3Mg_{}Cl_2+Cr_2O_3\rightarrow3MgO+2CrCl_3[/tex]

We now have the balanced equation. We have 3 Mg atoms, 6 Cl atoms, 2 Cr atoms, and 3 O atoms on each side of the reaction.

Consider the reaction between sodium metal and chlorine gas to form sodium chloride (table salt):2 Na(s) + Cl2 (g) > 2 NaCI(s)If 12.5 g of sodium react with sufficient chlorine, how many grams of sodium chloride should form?1. 12.1 grams2. 1.18 x 10^2 grams3. 15.9 grams4. 3.18 x 10^1 grams5. 51.0 grams

Answers

Answer:

[tex]4\text{ : 3.18 }\times\text{ 10}^1\text{ grams}[/tex]

Explanation:

Here, we want to calculate the mass of sodium chloride formed

Firstly, we need to get the number of moles of sodium that reacted

To get this, we divide the mass of sodium by its atomic mass

We have that as:

[tex]\frac{12.5}{23}\text{ = 0.5435 mole}[/tex]

Now, let us get the number of moles of chlorine

From the equation of reaction, 1 mole of sodium produced 1 mole of sodium chloride

Thus, 0.5435 mole of sodium will also produce 0.5435 mole of sodium chloride

To get the mass of sodium chloride formed, we multiply the above number of moles by the molar mass of sodium chloride

The molar mass of sodium chloride is 58.44 g/mol

Thus, the mass of sodium chloride formed will be:

[tex]58.44\text{ }\times\text{ 0.5435 = 31.8 g}[/tex]

7) What is the volume of the liquid in graduated cylinder A before the rockwas added?AYour answer8060BE10080★8 points

Answers

To read the measuring cylinder we first count the number of smaller divisions between the marked interval. In this case it is 10 intervals. The marked intervals are 60 and 80, therefore between these values we have 20. We will now divide the 20 by the 10 intervalswhich is equal to 2. Now each interval has a numerical value of 2. We read from the bottom of the meniscus. The meniscus is ttouching the 5th line and so we say 5 x 2=10, we add this to 60.

Answer: 60+10= 70,

7. A given sample of gas is held in a container with the volume of 6.02 L with a temperature of 59.5℃ at a pressure of 1.20 atm. What is the final pressure when the sample of gas is administered to a new volume of 10.0 L at 20.2℃?

Answers

The final pressure of the sample gas based on the new volume and temperature would be 0.245 atm.

Combined gas law

The combined gas law is expressed as:

[tex]\frac{p_1v_1}{t_1}[/tex] = [tex]\frac{p_2v_2}{t_2}[/tex]

Where

[tex]p_1[/tex]= initial pressure[tex]v_1[/tex]= initial volume[tex]t_1[/tex]= initial temperature[tex]p_2[/tex]= final pressure[tex]v_2[/tex]= final volume[tex]t_2[/tex]= final temperature

In this case, we were given all the variables except the final pressure, [tex]p_2[/tex]:

[tex]p_1[/tex]= 1.20 atm[tex]v_1[/tex]= 6.02 L[tex]t_1[/tex]= 59.5℃[tex]v_2[/tex]= 10.0 L[tex]t_2[/tex]= 20.2℃

Making p2 the subject of the formula in the combined gas equation, we have:

[tex]p_2[/tex]= [tex]p_1v_1t_2/t_1v_2[/tex]

Next, let's substitute the given variables:

[tex]p_2[/tex]= 1.2 x 6.02 x 20.2/59.5 x 10.0

     = 145.9248/595

     = 0.245 atm

In other words, the final pressure when the sample of gas is administered to a new volume of 10.0 L at 20.2℃ is 0.245 atm'

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The total number of calcium atoms in the expression 3 cal 2 shown in the equation 3CaCl 2 +2Na 3 PO 4 Ca(PO 4 ) 2 +6 NaCl is:

Answers

Let's write the balanced chemical equation:

[tex]3CaCl_2+2Na_3PO_4\to Ca_3(PO_4)_2+6NaCl.[/tex]

You can see that in the reactants and in the products we have the same number of atoms for each element. This is due to the matter of conservation law.

The total number of calcium atoms that are reacting and produced in the reaction is 3.

Of the elements calcium, Ca, beryllium, Be, barium, Ba, and strontium, Sr, arrange in order of decreasing atomic radii. Explain your answer using the terms electron and energy level.

Answers

Order these elements Ca, Be, Ba, and Sr:

If we take a look at the periodic table trends we will see that the atomic radii increases from left to right and from top to bottom.

So let's look for our elements:

Be: atomic number 4

Ba: atomic number 56

Ca: atomic number 20

Sr: atomic number 38

They are all Alkaline Earth Metals, they are in the same group.

So we said that the atomic radii increases from top to bottom. If we have to arrange them in order of decreasing atomic radii, we have to put first the largest one. The Barium is the one that is in a lower position, then Sr, then Ca and finally Be.

So the order of our problem is Ba, Sr, Ca and Be.

Which is the explanation?

As we move down in the periodic table the number of electrons increases. They all have 2 electrons in the outer shell. But let's look at the electronic configuration of two of them:

Ca:

Ba:

Determine the number of atoms of O in 7.23 moles of Ca(NO3)2.

Answers

Answer: 43.38 number of atoms

Explanation: If you look at this formula, then this one molecule contains six oxygen atoms. That means number of moles of oxygen atoms. That is six times the number of moles and of calcium nitrate, That is six, multiply 7.23, which comes out to be 43.38 moles of oxygen atom.

Calculate the frequency and the energy of blue light that has a wavelength of 423 nm (Note => 423 nm = 4.23 * 10^7 m)

Answers

According to the Planck's equation,the energy and frequency of blue light are 4.699×10[tex]^-19[/tex] J and 2.363×10⁶ m[tex]^-1[/tex] respectively.

What is Planck's equation?

Max Planck discovered the theory which stated that energy is transferred in the form of discrete packs which are called quanta and thus proposed an equation called the Planck's equation which relates energy and frequency of a photon and is given as, E=hcυ or in terms of wavelength it is ,E=hc/λ.

The equation makes use of a constant which is called the Planck's constant and it's value is 6.626×10[tex]^-34[/tex] Js.

In the given problem, energy is calculated as,E=6.626×10[tex]^-34[/tex]×3×10⁸/4.23×10[tex]^-7[/tex]=4.699×10[tex]^-19[/tex] J.

The frequency  of light is calculated as follows,4.699×10[tex]^-19[/tex]/19.878×10[tex]^-26[/tex]=2.363×10⁶ m[tex]^-1[/tex].

Thus , the energy and frequency of blue light are  4.699×10[tex]^-19[/tex] J and 2.363×10⁶ m[tex]^-1[/tex] respectively.

Learn more about Planck's equation,here:

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