The second volume : V₂= 0.922 L
<h3>
Further explanation
</h3><h3>Given
</h3>
7.03 Liters at 31 C and 111 Torr
Required
The second volume
Solution
T₁ = 31 + 273 = 304 K
P₁ = 111 torr = 0,146 atm
V₁ = 7.03 L
At STP :
P₂ = 1 atm
T₂ = 273 K
Use combine gas law :
P₁V₁/T₁ = P₂V₂/T₂
Input the value :
0.146 x 7.03 / 304 = 1 x V₂/273
V₂= 0.922 L
Answer:
5
Explanation:
Firstly, we convert what we have to percentage compositions.
There are two parts in the molecule, the sulphate part and the water part.
The percentage compositions is as follows:
Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%
The water part = 100 - 64 = 36%
Now, we divide the percentages by the molar masses.
For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol
For the H2O = 2(1) + 16 = 18g/mol
Now we divide the percentages by these masses
Sulphate = 64/160 = 0.4
Water = 36/18 = 2
The ratio is thus 0.4:2 = 1:5
Hence, there are 5 water molecules.
I think it would be penicillin as it saved many lives in its time and ours.
Answer:
A model is developed for predicting oxygen uptake, muscle blood flow, and blood chemistry changes under exercise conditions. In this model, the working muscle mass system is analyzed. The conservation of matter principle is applied to the oxygen in a unit mass of working muscle under transient exercise conditions. This principle is used to relate the inflow of oxygen carried with the blood to the outflow carried with blood, the rate of change of oxygen stored in the muscle myoglobin, and the uptake by the muscle. Standard blood chemistry relations are incorporated to evaluate venous levels of oxygen, pH, and carbon dioxide.
Explanation:
