Answer:
- 0.1852
- 0.0947
- 0.7201
- 3.0345 kg CO
/ Kg C
H
- 15.3848 Kg air / kg C H
Explanation:
Molar masses of each product are :
Butane = 58 kg /kmol
Oxygen = 32 kg/kmol
Nitrogen = 28 kg/kmol
water = 18 kg/kmol
<u><em>1) Calculate the mass fraction of carbon dioxide </em></u>
= ( 4 * 44 ) / ( (5 * 18) + (4 *44 )+ (24.44 * 28) )
= 176 / 950.32
= 0.1852
<em><u>2) Calculate the mass fraction of water </u></em>
= ( 5 * 18 ) / (( 5* 18 ) + ( 4*44) + ( 24.44 * 28 ))
= 90 / 950.32
= 0.0947
<em><u>3) Calculate the mass fraction of Nitrogen </u></em>
= (24.44 * 28 ) / ((4 * 44 ) + ( 24.44 * 28 ) + ( 5 * 18 ))
= 684.32 / 950.32
= 0.7201
<em><u>4) Calculate the mass of Carbon dioxide in the products</u></em>
Mco2 = ( 4 * 44 ) / 58 = 3.0345 kg CO / Kg C H
<u>5) Mass of Air required per unit of fuel mass burned </u>
Mair = ( 6.5 * 32 + 24.44 *28 ) / 58 = 15.3848 Kg air / kg C H
D. the activation energy is the displacement of energy needed for the reaction, so the distance underneath the maximum point
Answer:
[KCl] = 1.2 M
Explanation:
We need to complete the reaction:
2KCl(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCl₂(s)↓
By stoichiomety we know that 1 mol of chloride needs 1 mol of nitrate to react:
Let's find out the moles of nitrate, we have:
Molarity = mol/volume(L)
We convert the volume → 30 mL . 1L/1000mL = 0.030L
Molarity . volume(L) = moles → 0.400 M . 0.030L = 0.012 moles
Therefore, we can make a rule of three.
1 mol of nitrate reacts with 2 moles of chloride
Then, 0.012 moles of nitrate must react with (0.012 . 2) / 1 = 0.024 moles of KCl
We convert the volume from mL to L → 20 mL . 1L /1000mL = 0.020L
Molarity = mol /volume(L) → 0.024 mol /0.020L = 1.2 M
The answer to your question is Sea Breeze.
Answer: the atomic number 7 is
Nitrogen
Explanation:
