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olga_2 [115]
2 years ago
5

How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mm hg and a temperature of 21.0°c, can be made

from 1.566 g of zn and excess hcl? the partial pressure of water vapor is 18.65 mm hg at 21.0°c.
a.0.599 l?
Chemistry
1 answer:
natta225 [31]2 years ago
7 0
Answer:  
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.  
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn  
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced  
Now use the ideal gas law to find the volume V.  
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm  
T = 21 C + 273 = 294 K  
PV = nRT 
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L
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A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl
Alecsey [184]

Answer: 32.94 g

Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

Kr+2Cl_2\rightarrow KrCl_4

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

T = 350.8 + 273 = 623.8 K

For krypton, n=\frac{0.500*15.0}{0.0821*623.8}

n = 0.146 moles

for chlorine, n=\frac{1.50*15.0}{0.0821*623.8}

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})

= 32.94 g of KrCl_4

So, 32.94 g of the product will form.

5 0
2 years ago
Consider a saturated solution formed when 17.5 g of a solute dissolve in 28.3 g of a solvent, giving a total solution volume of
STALIN [3.7K]

Answer:

a) 38.2 % mass

b) 61.8 g solute/100 g solvent

c) 1.65 g/mL

Explanation:

Given the data:

mass of solute = 17.5 g

mass of solvent= 28.3 g

total solution volume= 27.8 mL

a)- mass percent= mass of solute/mass of solution x 100

mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g

mass % = 17.5 g/45.8 g x 100 = 38.2 % mass

b)- solubility = grams of solute/ 100 g solvent

                    = 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent  

c)- density = massof solution/total volumesolution  = 45.8 g/27.8 mL = 1.65 g/mL

7 0
2 years ago
Several amino acids are intermediates of the urea cycle, having side ammonia groups that join with free carbon dioxide (CO2) and
mr_godi [17]

Answer:

A. Arginine

Explanation:

The urea cycle is the cycle of the biochemical reactions which produces urea from ammonia.

Steps of the urea cycle:

  • Carbamoyl phosphate, in presence of ornithine transcarbamoylase is converted to citrulline by the denotation of carbamoyl phosphate groupto ornithine and a phosphate group is released.
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7 0
2 years ago
Consider the following reaction (X = Cl or Br) which statement s is are correct?
statuscvo [17]

A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll

5 0
2 years ago
Compound a on ozonolysis yields acetophenone and propanal. what is the structure of compound a? 1-phenyl-1-hexene 1-phenyl-2-pen
astra-53 [7]
Answer:
            2-Phenyl-2-Penetene on ozonolysis <span>yields acetophenone and propanal.

Explanation:
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6 0
2 years ago
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