Answer:
C. 101.2 L
Explanation:
N2 + H2= NH3
Balancing it,
N2 + 3 H2 = 2.NH3
(1 mol) (3 mol) (2 mol)
which means
1 molecule of nitrogen reacts with 3 molecule of hydrogen to form ammonia.
Likewise,
50.6 l of nitrogen reacts with 50.6 × 3= 151.8 l of hydrogrn to form 50.6 × 2= 101.2 l of ammonia.
To compute the percent error, you divide the absolute difference between the actual value and the measured value with the actual value. Then, multiply it with 100 to get the percentage. The solution is as follows:
Percent error = (250 - 243.9)/250 * 100
Percent error = 2.44%
Answer:
7.640 kg
Explanation:
Step 1: Write the balanced complete combustion equation for ethanol
C₂H₆O + 3 O₂ ⇒ 2 CO₂ + 3 H₂O
Step 2: Calculate the moles corresponding to 4 kg (4000 g) of C₂H₆O
The molar mass of C₂H₆O is 46.07 g/mol.
4000 g × 1 mol/46.07 g = 86.82 mol
Step 3: Calculate the moles of CO₂ released
86.82 mol C₂H₆O × 2 mol CO₂/1 mol C₂H₆O = 173.6 mol CO₂
Step 4: Calculate the mass corresponding to 173.6 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
173.6 mol × 44.01 g/mol = 7640 g = 7.640 kg
The answer for this is because 98,8
This question is describing two chemical equations whereby the concentration of ammonia has to be determined. The first reaction is between 25.00 mL of ammonia and 50.00 mL of 0.100-M HCl whose excess was neutralized with 21.50 mL of 0.050-M Na₂CO₃ and thus, the concentration ammonia in the cloudy solution was determined as 0.114 M.
First of all we need to go over the titration of the excess HCl with Na₂CO₃ by writing the chemical equation it takes place when they react:

Whereas the mole ratio of HCl to Na₂CO₃ is 2:1 and the volume of the HCl leftover is determined as follows:

Next, we infer that the consumed volume of HCl by the ammonia solution was:

Then, we write the chemical equation that takes place between ammonia and HCl:

Whereas the mole ratio is now 1:1, which means that the concentration of ammonia was:

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