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3 years ago

11. Calculate the energy required to produce 12.00 mol of CI207 on the basis of the

Chemistry

1 answer:

Ber [7]3 years ago

4 0

Answer: 780 kcal are required to produce 12.00 mol of $Cl_2O_7$

Explanation:

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and written along with reactants.

The balanced chemical reaction is:

$2Cl_2(g)+7O_2(g)+130kcal\rightarrow 2Cl_2O_7(g)$

According to stoichiometry :

2 moles of $Cl_2O_7$ are produced by absorption of energy = 130 kcal

Thus 12.00 moles of $Cl_2O_7$ are produced by absorption of energy = $\frac{130}{2}\times 12.00=780kcal$

Thus 780 kcal are required to produce 12.00 mol of $Cl_2O_7$

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How much water would need to be added to 1092 mL of a 54.7 M NaCl solution to make a 0.25 M solution?

babunello [35]

Answer:

237.8L of water would need to be added.

Explanation:

The first thing to do is to identify that the equation to be used is M1V1=M2V2. (This equation works because it turns everything into moles which can then be compared).

Then figure out what information you have and what is being found. In this case:

M1 = 54.7 M

V1 = 1092 mL = 1.092 L

M2 = 0.25 M

V2 = unknown

Then solve the equation for whatever you are trying to find.

M1V1=M2V2

V2=M1V1/M2

Now you need to plug everything in.

V2=(54.7M*1.091L)/0.25M

V2=238.93L

That means that the solution needs a volume of 238.7L to gain a molarity of 0.25M but the starting solution already had a volume of 1.092 L meaning that to find the amount of solvent that needs to be added you just subtract the starting volume by the volume that the solution needs to be.

238.93L - 1.091L = 237.8L

Therefore the answer is that 237.8L needs to be added to a 1.092L 54.7M NaCl solution to make the concentration 0.25M.

I hope this helps. Let me know if anything is unclear.

8 0

3 years ago

Electron transfer is the term for this process. The resulting anion and cation are attracted by Coulombic forces and an ionic co

Nitella [24]

<h2>The compound formed is NaCl and $CaCl_{2}$ </h2>

Explanation:

PART 1:

Atomic number of sodium is 11.
Sodium(Na, Z = 11, Group 1A) will lose 1 electron to become $Na^{1+}$ which is isoelectronic to Neon (Ne ,Z = 10).
The symbol for the sodium ion is $Na^{+}$ .
Chlorine atom ( $Cl$ , Z = 17, Group 7A or 17) gains 1 elecron to be isoelectronic to neon.
The symbol for the compound formed is NaCl.
PART 2:
Atomic number of calcium is 20
Calcium(Ca, Z = 20 ,Group 2A) will lose two electrons to become $Ca^{2+}$ which is isoelectronic to argon.
The symbol for the ion is $Ca^{2+}.$
Two chlorine atoms (Cl, Z = 17, Group 7A or 17) each gains one electron to be isoelectronic to argon(Z = 18)
The symbol for the compound formed is $CaCl_{2}$ .

8 0

3 years ago

PLEASE HELPPPP!!!!!

Likurg_2 [28]

Answer : The mass of nitric acid is, 214.234 grams.

Solution : Given,

Moles of nitric acid = 3.4 moles

Molar mass of nitric acid = 63.01 g/mole

Formula used :

$\text{Mass of }HNO_3=\text{Moles of }HNO_3\times \text{Molar mass of }HNO_3$

Now put all the given values in this formula, we get the mass of nitric acid.

$\text{Mass of }HNO_3=(3.4moles)\times (63..01g/mole)=214.234g$

Therefore, the mass of nitric acid is, 214.234 grams.

6 0

3 years ago

PLEASE HELP!! I REALLY NEED HELP

Allushta [10]

Answer:

Explanation:

1. find the molar mass (amu) of each element and add them to get the whole molar mass.

2. divide the 1 element molar mass with the whole molar mass

3. multiple by 100 and that gives you the % composition.

1. Na(22.99amu) + Cl (35.453amu)=58.443

2(Na): $\frac{22.99}{58.443}$ = .393

2(Cl): $\frac{35.453}{58.443}$ = .607

3(Na): .393 * 100=39.3%

3(Cl): .607 * 100= 60.7%

1. K: (39.098)(2)=78.196

_ C: (12.011)(1)= 12.011

_O: (15.99)(3) = 47.997

78.196+12.011+47.997= 138.204

2:K: $\frac{78.196}{138.204}$ = .566 <u>Step </u>3: (.566)(100)= 56.6%

2: C: $\frac{12.011}{138.204}$ = .087 <u>Step 3</u>: (.087)(100)= 8.7%

2: O: $\frac{47.997}{138.204}$ = .347 <u>Step 3</u>: (.347)(100) = 34.7%

1. Fe (55.845)(3)= 167.535

_ O (15.999)(4) = 63.996

167.535+63.996=231.531

2: Fe: $\frac{167.535}{231.531}$ = .724 Step 3: (.724)(100)= 72.4%

2: O : $\frac{63.996}{231.531}$ = .276 Step 3: (.276)(100) = 27.6%

C(12.011*3)=36.033

H(1.008*5)=5.04 + (1.008*3)=3.024 so its 8.064

O(15.999*3)=47.997

add them: 92.094

2: C: $\frac{36.033}{92.094}$ = .391 Step 3: (.391)(100) = 39.1%

2: H: $\frac{8.064}{92.094}$ = .088 step 3: (.088)(100) = 8.8%

2: O: $\frac{47.997}{92.094}$ = .521 step 3: (.521)(100) = 52.1%

3 0

3 years ago

What is the percent by mass of carbon in C10 H14 N2?

MrMuchimi

Multiply the number of each element by its mass on the periodic table.

Add them together.

Put the mass of carbon only over the total mass and multiply the result by 100 to get the percentage.

8 0

3 years ago

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