Answer:
d
Explanation:
they would all have the same volumes
Answer:
83ºC
Explanation:
A bomb calorimeter is an instrument used to measure the heat that release or absorb a particular reaction.
The reaction of combustion of propane is:
C₃H₈ + 5O₂ → 3 CO₂ + 4 H₂O ΔH = -2222kJ/mol
<em>1 mole of propane release 2222kJ</em>
10.0g of propane (Molar mass: 44.1g/mol).
10.0g ₓ (1mol/ 44.1g) = <em>0.227 moles of C₃H₈</em>
If 1 mole of propane release 2222kJ, 0.227moles will release (Release because molar heat is < 0):
0.227 moles of C₃H₈ ₓ (2222kJ / mol) = 504kJ.
Our calorimeter has a constant of 8.0kJ/ºC, that means if there are released 8.0kJ, the bomb calorimeter will increase its temperature in 1ºC. As there are released 504kJ:
504kJ ₓ (1ºC / 8.0kJ) = 63ºC will increase the temperature in the bomb calorimeter.
As initial temperature was 20ºC, final temperature will be:
<h2>83ºC</h2>
Answer is: <span>b) 9,042 L.
Ideal gas law: p</span>·V = n·R·T.
p is the pressure of the gas.
V is the volume of the gas.
n is amount of substance.
R is universal gas constant.
T is temperature.
T₁ = 15°C = 288 K.
V₁ = 10,5 L.
T₂ = -25°C = 248 K.
V₂ = ?
V₁/T₁ = V₂/T₂
V₂ = 10,5 L · 248 K ÷ 288 K
V₂ = 9,042 L.
The balanced equation is Mg + 2AgNO₃ ⟶ Mg(NO₃)₂ + 2Ag
Step 1. Write the <em>unbalanced equation
</em>
Mg + AgNO₃ ⟶ Mg(NO₃)₂ + Ag
Step 2. Start with the<em> most complicated-looking formula</em> [Mg(NO₃)₂] and balance its atoms.
Mg: Already balanced —1 atom each side.
N: We need 2 N on the left. Put a 2 in front of AgNO₃.
1Mg + 2AgNO₃ ⟶ 1Mg(NO₃)₂ + Ag
O: Already balanced —6 atom6 each side.
Step 3: Balance <em>Ag</em>
We have 2Ag on the left. We need 2Ag on the right.
1Mg + 2AgNO₃ ⟶ 1Mg(NO₃)₂ + 2Ag
Answer:
b) pH = 9.25
Explanation:
- NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
- NH3 + H2O ↔ NH4+ + OH-
- 2 H2O ↔ H3O+ + OH-
⇒ Kb = [ NH4+ ] * [ OH- ] / [ NH3 ] = 1.86 E-5......from literature
mass balance NH4+:
⇒ M NH4+ = [ NH4+ ] - [ OH- ]
∴ [ NH3 ] ≅ M NH4+ = 0.26 M
⇒ Kb = (( 0.26 + [ OH- ] )) * [ OH- ] / 0.26 = 1.86 E-5
⇒ 0.26 [ OH-] + [ OH- ]² = 4.836 E-6
⇒ [ OH- ]² + 0.26 [ OH- ] - 4.836 E-6 = 0
⇒ [ OH- ] = 1.859 E-5 M
⇒ pOH = - Log ( 1.859 E-5 )
⇒ pOH = 4.7305
⇒ pH = 14 - pOH = 9.269
