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Galina-37 [17]
3 years ago
10

Use the information to answer the following question.

Chemistry
2 answers:
d1i1m1o1n [39]3 years ago
6 0

Answer: D

Explanation:

jolli1 [7]3 years ago
4 0
I think the answer is c
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How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below?
ankoles [38]
The chemical reaction equation for this is 

XeF6 + 3H2 ---> Xe + 6HF

Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.

we get n = 0.1433 moles H2

to get the mass of XeF6, 

we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.

0.1433 moles H2 x \frac{1 mole XeF6}{3 moles H2} x \frac{245.28 g XeF6}{1 mole XeF6} = 11.71 g XeF6

Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
3 0
3 years ago
The number of protons plus the number of neutrons in an atom is called its 7 number. for carbon-12, this number is
emmasim [6.3K]
The mass of Carbon-12 is 12.
8 0
3 years ago
What is the atomic mass of copper?
kipiarov [429]

Answer:

63.546 u

Explanation:

4 0
3 years ago
4.80 x 10^25 formula units pf calcium iodide (Cal2) will have what mass?
maw [93]

Answer:

mass CaI2 = 23.424 Kg

Explanation:

From the periodic table we obtain for CaI2:

⇒ molecular mass CaI2: 40.078  + ((2)(126.90)) = 293.878 g/mol

∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708  mol CaI2

⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g

⇒ mass CaI2 = 23.424 Kg

7 0
3 years ago
Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 8 moles of Al react with 8 moles of HBr, what is the limiting reactant?
TiliK225 [7]

Answer:- HBr is limiting reactant.

Solution:- The given balanced equation is:

2Al+6HBr\rightarrow 2AlBr_3+3H_2

From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.

The calculations could be shown as:

8molAl(\frac{6molHBr}{2molAl})

= 24 mol HBr

From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.

7 0
3 years ago
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