Determine the specific heat capacity of an alloy that requires 59.3 kj to raise the temperature of 150.0 g alloy from 298 k to 3
1 answer:
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Answer: a) 211 mm Hg
b) 0.629 grams
Explanation:
According to Dalton's law, the total pressure is the sum of individual pressures.
= total pressure = 750 mmHg
= 124 mm Hg
= 218 mm Hg
= 197 mm Hg
= ?
Thus the partial pressure of the helium gas is 211 mmHg.
b) According to the ideal gas equation:
P = Pressure of the gas = 211 mmHg = 0.28 atm (760mmHg=1atm)
V= Volume of the gas = 13.0 L
T= Temperature of the gas = 282 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= ?
Mass of helium=
Thus mass of helium gas present in a 13.0-L sample of this mixture at 282 K is 0.629 grams
Considering the reaction stoichiometry and Avogadro's Number, the mass of barium chloride produced is 21.6 grams.
<h2>Balanced reaction</h2>
The balanced reaction is
2 HCl + Ba → BaCl₂ + H₂
<h2>Moles of HCl that react</h2>Avogadro's Number is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.
Then you can apply the following rule of three: if 6.023×10²³ molecules are contained in 1 mole of HCl, then 1.25×10²³ molecules are contained in how many moles of HCl?
amount of moles of HCl= (1.25×10²³ molecules × 1 mole)÷ 6.023×10²³ atoms
<u><em>amount of moles of HCl= 0.2075 moles</em></u>
Then, 0.2075 moles of HCl react.
<h2>Reaction stoichiometry</h2>By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- HCl: 2 moles
- Ba: 1 mole
- BaCl₂: 1 mole
- H₂: 1 mole
Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl produce 1 mole of BaCl₂, 0.2075 moles of HCl will produce how many moles of BaCl₂?
<u><em>amount of moles of BaCl₂= 0.10375 moles</em></u>
Being the molar mass of BaCl₂ 208.24 g/mole, then the mass of barium chloride produced is calculated as:
Finally, the mass of barium chloride produced is 21.6 grams.
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