(60)/(60+5.05)=.922367 C
1-0.922367=0.07763259 H
(0.922367)(78.12)=72.05534204 C
(0.07763259)(78.12)=6.06 H
72.05534204/(12.01)=6 C
6.06/1.01=6 H
Empirical= CH
Molecular=C6H6
Albinism is a recessive trait.
Na2S2O3(aq) + 4Cl2( g) + 5H2O = 2NaHSO4(aq) + 8HCl(aq)
1)How many moles of HCl can form from 0.21 mol of Cl2?
0.21 mol Cl2 ( 8 mol HCl / 4 mol Cl2 ) = 0.42 mol HCl
2)How many moles of H2O are required for the reaction of 0.18 mol of Cl2?
0.18 mol Cl2 ( 5 mol H2O / 4 mol Cl2 ) = 0.225 mol H2O
3)How many moles of H2O react if 0.50 mol HCl is formed?
0.50 mol HCl ( 5 mol H2O / 8 mol HCl ) = 0.3125 mol H2O
Answer:first one: 3.0g second one: 10g H2O(I)
Explanation:
Answer:
43.75 g of Nitrogen
Explanation:
We'll begin by calculating the mass of 1 mole of NH₄NO₃. This can be obtained as follow:
Mole of NH₄NO₃ = 1 mole
Molar mass of NH₄NO₃ = 14 + (4×1) + 14 + (3×16)
= 14 + 4 + 14 + 48
= 80 g/mol
Mass of NH₄NO₃ =?
Mass = mole × molar mass
Mass of NH₄NO₃ = 1 × 80 = 80 g
Next, we shall determine the mass of N in 1 mole of NH₄NO₃.
Mass of N in NH₄NO₃ = 2N
= 2 × 14
= 28 g
Thus,
80 g of NH₄NO₃ contains 28 g of N.
Finally, we shall determine the mass of N in 125 g of NH₄NO₃. This can be obtained as follow:
80 g of NH₄NO₃ contains 28 g of N.
Therefore, 125 g of NH₄NO₃ will contain = (125 × 28) / 80 = 43.75 g of N.
Thus, 125 g of NH₄NO₃ contains 43.75 g of Nitrogen
