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Irina18 [472]
3 years ago
5

What volume (in L) of H2 would be collected at 22.0oC and a pressure of 713 torr if 2.65 g of zinc react according to the equati

on:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Please provide the steps and explanations. I truly do not know where to start with this question.
Thanks
Chemistry
1 answer:
ASHA 777 [7]3 years ago
4 0

Answer: 83.74L

Explanation:

Temp. = 295K

P = 713torr = 0.938atm

Mass = 2.65g

PV = nRT

V = nRT/PV

n = Mass/Molar mass

Molar mass of Hydrogen gas = 1.00784*2= 2.0156g/mok

n = 2.65/2.0156 = 1.31469mol

V = 1.31469*0.08205*295/0.938

V = 83.74L

The volume = 83.74L

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Complete combustion of 5.90 g of a hydrocarbon produced 18.8 g of CO2 and 6.75 g of H2O.
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First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C 
Molar mass of H2O:
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As there is 2 moles of H in H2O,
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<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>

<span>Divide both number of moles by the smaller number of moles: </span>
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So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>

<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula: 
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3 years ago
Three samples of the same metal are weighed and their masses are found to be 44.40 g, 40.58 g, and 38.35 g. The corresponding vo
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The density of a material is the mass of the material per unit volume. Here the weight of the same metal is 44.40g, 40.58g and 38.35g having volume 4.8 mL, 4.7 mL and 4.2 mL respectively. Thus the density of the metal as per the given data are, \frac{44.40}{4.8} = 9.25g/mL, \frac{40.58}{4.7} = 8.634g/mL and \frac{38.35}{4.2} = 9.130g/mL respectively.

The equation of the standard deviation is √{∑(x  - \frac{}{x})÷N}

Now the mean of the density is {(9.25 + 8.634 + 9.130)/3} = 9.004 g/mL.

The difference of the density of the 1st metal sample (9.25-9.004) = 0.246 g/mL. Squaring the value = 0.060.

The difference of the density of the 2nd metal sample (9.004-8.634) =0.37 g/mL. Squaring the value = 0.136.

The difference of the density of the 3rd metal sample (9.130-9.004) = 0.126 g/mL. Squaring the value 0.015.

The total value of the squared digits = (0.060 + 0.136 + 0.015) = 0.211. By dividing the digit by 3 we get, 0.070. The standard deviation will be \sqrt{0.070}=0.265. Thus the standard deviation of the density value is 0.265g/mL.  

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