Answer:
a. 92.4%
Explanation:
Based on the reaction:
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)
To obtain the percent yield you need to obtain moles of trona and calculate thoeretical moles of Na₂CO₃, and the ratio of obtained moles / theoretical moles of Na₂CO₃ give percent yield, thus:
Moles of trona:
1.00 metric ton × (1x10³kg / 1 metric ton) × ( 1000moles /226.03 kg) = <em>4424 moles</em>
The theoretical moles of Na₂CO₃ that produce 4424 moles of trona are (Based on the reaction, 2 moles of trona produce 3 moles of Na₂CO₃):
4424 moles trona × (3 moles Na₂CO₃ / 2 moles trona) = <em>6636 moles of Na₂CO₃.</em>
The obtained moles of Na₂CO₃:
0.650 metric ton × (1x10³kg / 1 metric ton) × (1000 moles / 105.99kg) = <em>6133 moles</em>
The ratio of obtained moles / theoretical moles gives:
6133 moles / 6636 moles = 0.924 = <em>92.4%</em>
I hope it helps!
Conditions:
Low pressure and low temperature
Low pressure and high temperature
High pressure and low temperature
High pressure and high temperature
Answer:
N2
Explanation:
Rate of effusion is defined by Graham's Law:
(Rate 1/Rate 2) = (sqrt (M2)/ sqrt (M1))
(Where M is the molar mass of each substance. )
Molar Mass of oxygen, O2, is 32 (M1).
Rate of effusion of O2 to an unknown gas is .935(Rate 1).
Rate 2 is unknown so put 1.
Solve for x (M2).
.935/1 = sqrt x/ sqrt32
.935 x sqrt 32 = sqrt x
5.29 = sq rt x
5.29^2 = 27.975 = 28
N2 has a molar mass of 28 so it is the correct gas.
The compound : C₄₀H₄₄N₄O
<h3>Further explanation</h3>
The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components
The principle of determining empirical formula
• Determine the mass ratio of the constituent elements of the compound.
• Determine the mole ratio by dividing the percentage by the atomic mass
The mol ratio of composition : C : H : N : O
Explanation:
The answer to questions are
A) 4
B) 3
C) 5
D) 3
E) 3
