Answer:
The volume inside the balloon is = 121 
Explanation:
Temperature T = - 1 °c = 272 K
Pressure = 0.7 atm = 71 k pa
No. of moles = 3.8
Mass of the gas inside the volume = 3.8 × 4 = 15.2 kg
From ideal gas equation
P V = m R T
Put all the values in above formula we get
71 × V =15.2 × 2.077 × 272
V = 121 
Therefore the volume inside the balloon is = 121 
C is the answer 3 + 2 +8 = 13
Ribosome would be considered a factory “worker” as they *produce* proteins!
Answer:
39.6 g
Explanation:
The equation of the reaction is;
2Mg(s) + O2(g) --------> 2MgO(s)
To obtain the limiting reactant;
Number of moles in 26.4 g of Mg = 26.4g/24 g/mol = 1.1 moles
If 2 moles of Mg yields 2 moles of MgO
1.1 moles of Mg yields 1.1 * 2/2 = 1.1 moles of MgO
Number of moles in 26.4 g of O2 = 26.4 g/32g/mol = 0.825 moles
If 1 mole of O2 yields 2 moles of MgO
0.825 moles of O2 yields 0.825 moles * 2/1 = 1.65 moles of MgO
Hence Mg is the limiting reactant.
Theoretical yield of MgO = 1.1 moles of MgO * 40 g/mol = 44 g
Percent yield = 90%
Percent yield = actual yield/theoretical yield * 100
Actual yield = Percent yield * theoretical yield/100
Actual yield = 90 * 44/100
Actual yield = 39.6 g
Lead Nitrate is highly soluble in water. 37.65g/100 mL at 20*C