The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
Molar mass Ra(OH)2 = 260 g/mol
<span>Moles = 987 g / 260 = 3.80 moles</span>
Answer:
The answer to your question is: 0.028 kg of NO2
Explanation:
Data
3.7 x 10²⁰ molecules of NO2 in kg
MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg
1 mol of NO2 --------------------- 6.023 x 10 ²³ molecules
x --------------------- 3.7 x 10²⁰ molecules
x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³
x = 0.00061 mol
1 mol of NO2 --------------------- 46 kg of NO2
0.00061 mol ------------------ x
x = 0.00061 x 46/1
x = 0.028 kg of NO2
