Answer: Thus 0.724 mol of 
are needed to obtain 18.6 g of 
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of are produced by = 1 mole of
Thus 1.09 moles of will be produced by =
of
But as yield of reaction is 75.6 %, the amount of needed is =
Thus 0.724 mol of are needed to obtain 18.6 g of
Answer:
B. Green solution density is 1.06 g/ml and blue solution density is 1.20 g/ml
Explanation:
Density is given as,
D = Mass / Volume
Red Solution,
D = 25 g / 25 mL
D = 1 g/mL
Green Solution,
D = 26.5 g / 25 mL
D = 1.06 g/mL
Yellow Solution,
D = 28.2 g / 25 mL
D = 1.128 g/mL
Blue Solution,
D = 30 g / 25 mL
D = 1.20 g/mL
The temperature increase is from 19.5 to 100 degrees centigrade or 80.5 degrees centigrade. The calorie increase is 2.50 x 1000 x 0.238902957619 or a total of 597.25 calories. 597.25/80.5 = 7.419 calories per degree centigrade. 7.419/135 grams = 0.0549 calories/gram/degree centigrade. The conversion from kilo joules involves multiplying the calories per joule x 1000 to get the number of calories in one kilo joule and then by the 2.5.
Explanation:
We have to find the number of moles of N₂ that are present in a sample that has a volume of 40.0 L at STP.
STP means Standard Conditions of Temperature and Pressure. These conditions are 273.15 K and 1 atm. We know that 1 mol of N₂ will occupy 22.4 L. We can use that ratio to find the answer to our problem.
1 mol of N₂ = 22.4 L
moles of N₂ = 40.0 L * 1 mol/(22.4 L)
moles of N₂ = 1.79 mol
Answer: 1.79 moles of nitrogen are present.
<h3>
Answer:</h3>
134 atm
<h3>
Explanation:</h3>
- Based on the pressure law, the pressure of a gas varies directly proportionally to the absolute temperature at a constant volume.
- Therefore; we are going to use the equation;
In this case;
Initial pressure, P1 = 144 atm
Initial temperature, T1 (48°C) = 321 K
Final temperature, T2 (25°C) = 298 K
We need to find the final pressure,
Therefore;
P2 = (P1/T1)T2
= (144/321)× 298 K
= 133.68 atm
= 134 atm
Therefore, the new pressure will be 134 atm.
