Answer:
There are 2 nitrogen atoms on the product side.
N2 + 3H2
There are 6 hydrogen atoms on the reactant side.
2NH3
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The volume occupied by 0.102 mole of the helium gas is 2.69 L
<h3>Data obtained from the question</h3>
The following data were obtained from the question:
- Number of mole (n) = 0.102 moles
- Pressure (P) = 0.95 atm
- Temperature (T) = 305 K
- Gas constant (R) = 0.0821 atm.L/Kmol
- Volume (V) =?
<h3>How to determine the volume </h3>
The volume of the gas can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
Divide both sides by P
V = nRT / P
V = (0.102 × 0.0821 × 305) / 0.95
V = 2.69 L
Thus, the volume of the gas is 2.69 L
Learn more about ideal gas equation:
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Answer:
The answer to your question is: 0.3 moles of AgNO₃
Explanation:
1.0 L sample
0.1 mol of NaCl
0.1 mol of CaCl₂
AgNO₃ = ? moles
Reactions
NaCl + AgNO₃ ⇒ AgCl + NaNO₃
Then 1 NaCl mol --------------- 1 AgNO₃
0.1 mol -------------- x
x = 0.1 moles of AgNO₃ needed
CaCl₂ + 2 AgNO₃ ⇒ 2 AgCl + Ca(NO₃)₂
Then 1 mol of CaCl₂ ------------- 2 moles of AgNO₃
0.1 mol ------------- x
x = 0.2 moles of AgNO₃
Total moles of AgNO₃ = 0.1 + 0.2 = 0.3
True
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