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PtichkaEL [24]
2 years ago
10

Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at

689 torr and 39 ∘C, is required to form 28 g of NaCl?
Chemistry
1 answer:
asambeis [7]2 years ago
6 0

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

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If 2.0 ml of 6.0m hcl is used to make a 500.0-ml aqueous solution, what is the molarity of the dilute solution?
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Naturally occurring iodine has an atomic mass of 126.9045 amu. A 12.3849-g sample of iodine is accidentally contaminated with 1.
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Answer:

127.0665 amu

Explanation:

Firstly, to answer the question correctly, we need to access the percentage compositions of the iodine and the contaminant iodine. We can do this by placing their individual masses over the total and multiplying by 100%.

We do this as follows. Since the mass of the contaminant iodine is 1.00070g, the mass of the 129I in that particular sample will be 12.3849 - 1.00070 = 11.3842g

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[8.1/100 * 128.9050] + [91.9/100 * 126.9045] = 10.441305 + 116.6252355 = 127.0665 amu

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