Answer:
There will be 143,67g CO2 produced
Explanation:
2 C6H6 + 15 O2 → 12 CO2 + 6 H2O
(42,5 g C6H6) / (78.1124 g C6H6/mol) = 0.54408775 mole C6H6
(113.1 g O2) / (31.9989 g O2/mol) = 3.534496 moles O2
0.54408775 mole of C6H6 would react completely with 0.54408775 x (15/2) = 4.080658 mole O2, but there is more O2 present than that, so O2 is in excess and C6H6 is the limiting reactant.
(0.54408775 mol C6H6) x (12/2) x (44.0096 g/mol) = 143.67 g CO2
The rate of effusion of gas B is 728.32 m/s
<u>Explanation:</u>
Given:
Mass of A, m₁ = 46 g/mol
Rate of effusion of A, R₁ = 515 m/s
Mass of B, m₂ = 92 g/mol
Rate of effusion of B, R₂ = ?
We know:
Substituting the value we get:
Therefore, the rate of effusion of gas B is 728.32 m/s
Answer:
1. Balanced chemical for the given chemical reaction is :
number of mole of sulphuric acid produced = 12.5 mol
number of mole of oxygen required=6.25 mol
Explanation:
1. Balanced chemical for the given chemical reaction is :
from above balanced equation it is clearly that ,
2 mole of sulpher dioxide gives 2 mole of sulphuric acid
1 mole of sulpher dioxide gives 1 mole of sulphuric acid
therefore,
12 .5 mole of sulpher dioxide will givs 12.5 mole of sulphuric acid
number of mole of sulphuric acid produced = 12.5 mol
2 mole of sulphur dioxide needs 1 mole of oxygen gas
so,
1 mole of sulphur dioxide needs 0.5 mole of oxygen gas
therefore 12.5 of sulphur dioxide needs mole of oxygen
number of mole of oxygen required=6.25 mol
Answer:
0.05142 M
Explanation:
Let's consider the neutralization reaction between KHP and NaOH.
KHC₈H₄O₄ + NaOH → H₂O + NaKC₈H₄O₄
12.39 mL of 0.04150 M NaOH were required for the neutralization. The moles that reacted are:
12.39 × 10⁻³ L × 0.04150 mol/L = 5.142 × 10⁻⁴ mol
The molar ratio of KHP to NaOH is 1:1. The moles of KHP that reacted are 5.142 × 10⁻⁴ mol.
5.142 × 10⁻⁴ moles of KHP were in 10 mL of solution. The concentration of the KHP solution is:
5.142 × 10⁻⁴ mol / 10 × 10⁻³ L = 0.05142 M
Answer:
the mass is m ' a so yes a and m
Explanation: first off ' m