Answer:
V = 34.55 L
Explanation:
Given that,
No of moles, n = 1.4
Temperature, T = 20°C = 20 + 273 = 293 K
Pressure, P = 0.974 atm
We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :
PV=nRT
R is gas constant, 
Finding for V,
So, the volume of the gas is 34.55 L.
Answer:
The answer to your question is: % error = 0.4
Explanation:
Data
real value = 22.48%
estimated value = 22.57 %
Formula
% error = |real value - estimated value|/real value x 100
%error = |22.48 - 22.57|/22.48 x 100
% error = |-0.09|/22.48 x 100
%error = 0.09/22.48 x 100
% error = 0.004 x 100
% error = 0.4
Answer:
2.1056L or 2105.6mL
Explanation:
We'll begin by calculating the number of mole in 10g of Na2CO3. This can be obtained as follow:
Molar mass of Na2CO3 = (23x2) + 12 + (16x3) = 106g/mol
Mass of Na2CO3 = 10g
Mole of Na2CO3 =.?
Mole = mass /molar mass
Mole of Na2CO3 = 10/106
Mole of Na2CO3 = 0.094 mole
Next, we shall determine the number of mole CO2 produced by the reaction of 0.094 mole of Na2CO3. This is illustrated below:
Na2CO3 + 2HCl —> 2NaCl + H2O + CO2
From the balanced equation above,
1 mole of Na2CO3 reacted to produce 1 mole of CO2.
Therefore, 0.094 mole of Na2CO3 will also react to 0.094 mole of CO2.
Next, we shall determine the volume occupied by 0.094 mole of CO2 at STP. This is illustrated below:
1 mole of a gas occupy 22.4L at STP. This implies that 1 mole CO2 occupies 22.4L at STP.
Now, if 1 mole of CO2 occupy 22.4L at STP, then, 0.094 mole of CO2 will occupy = 0.094 x 22.4 = 2.1056L
Therefore, the volume of CO2 produced is 2.1056L or 2105.6mL
Answer:
T₁ = 39 K
Explanation:
Given data:
Initial pressure = 1023.6 kpa
Final pressure = 8114 kpa
Final temperature = 36°C (36+ 273= 309K)
Initial temperature = ?
Solution:
P₁/T₁ = P₂/T₂
T₁ = P₁×T₂ /P₂
T₁ = 1023.6 kpa × 309 K /8114 kpa
T₁ = 316292.4 K. Kpa /8114 kpa
T₁ = 39 K
Thus original pressure was 39 k.
Answer:
15.2 g H2
Explanation:
2H2O -> 2H2 + O2
9.06 x 10^24 molecules x (1 mol/6.022 x 10^23 molecules) x (2 mol H2/2 mol H2O) x (1.008 g/1 mol) = 15.2 g H2
