The empirical formula :
C₁₀H₁₆N₄SO₇
<h3>Further explanation</h3>
Given
6.4 g sample
Required
The empirical formula
Solution
mass C :
= 12/44 x 8.37 g
= 2.28
mass H :
= 2/18 x 2.75 g
= 0.305
mass N = 1.06
mass S :
= 32/64 x 1.23
= 0.615
mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g
Mol ratio :
= C : H : N : S : O
= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16
= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019
= 10 : 16 : 4 : 1 : 7
The empirical formula :
C₁₀H₁₆N₄SO₇
<span>Catalysts decrease the activation energy and the more collisions result in a </span>reaction<span>, so the </span>rate<span> of </span><span>reaction increases.</span><span />
Explanation:
Yes...the molecules would get warmer has they collide into each Other...
The action is interaction
The volume (in liters) of CO₂ that can be consumed at STP by 435 g Na₂O₂ is 125 L of Co₂
<u><em>calculation</em></u>
2Na₂O₂(s) +2 CO₂ (g)→ 2 Na₂CO₃(s) + O₂(g)
Step 1 : find the moles of Na₂O₂
moles = mass÷ molar mass
from periodic table the molar mass of Na₂O₂ = (23 x2) +( 16 x2) = 78 g/mol
moles= 435 g÷ 78 g/mol = 5.58 moles
Step 2: use the mole ratio to determine the moles of CO₂
from given equation Na₂O₂ : CO₂ =2 :2 =1:1
Therefore the moles of CO₂ is also = 5.58 moles
Step 3: find the volume of CO₂ at STP
that is at STP 1 mole of a gas = 22.4 L
5.58 moles = ? l
<em>by cross multiplication</em>
= (5.58 moles x 22.4 L) / 1 mole = 125 L
