The answer is Cl (chlorine)
Answer:
A = 674.33mmHg
B = 0.385atm
Explanation:
Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.
Mathematically,
P = kT, k = P / T
P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn
A)
Data:
P1 = 799mmHg
T1 = 50°C = (50 + 273.15) = 323.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2
P2 = (P1 × T2) / T1
P2 = (799 × 273.15) / 323.15
P2 = 674.37mmHg
The final pressure is 674.37mmHg
B)
P1 = 0.470atm
T1 = 60°C = (60 + 273.15)K = 333.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2,
P2 = (P1 × T2) / T1
P2 = (0.470 × 273.15) / 333.15
P2 = 0.385atm
The final pressure is 0.385atm
Answer:
The answer is 42, 39 grams of LiCl
Explanation:
We calculate the weight of 1 mol of LiCl, from the atomic weights of each element obtained from the periodic table:
Weight 1 mol LiCl= Weight Li + Weight Cl =6,94 g+ 35, 45 g= 42, 39grams
Answer:
290.82g
Explanation:
The equation for the reaction is given below:
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:
Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol
Mass of H2SO4 from the balanced equation = 3 x 98 = 294g
Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96] = 54 + 288 = 342g
Now, we can obtain the mass of aluminium sulphate formed by doing the following:
From the equation above:
294g of H2SO4 produced 342g of Al2(SO4)3.
Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3
Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.
