34% I believe
Hope this helps!
STSN
Answer:
0.0905 M
Explanation:
Let's consider the neutralization reaction between H2SO4 and KOH.
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:
0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol
1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:
M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M
Answer:
30.3 g
Explanation:
At STP, 1 mol of any gas will occupy 22.4 L.
With the information above in mind, we <u>calculate how many moles are there in 32.0 L</u>:
- 32.0 L ÷ 22.4 L/mol = 1.43 mol
Then we <u>calculate how many moles would there be in 16.6 L</u>:
- 16.6 L ÷ 22.4 L/mol = 0.741 mol
The <u>difference in moles is</u>:
- 1.43 mol - 0.741 mol = 0.689 mol
Finally we <u>convert 0.689 moles of CO₂ into grams</u>, using its <em>molar mass</em>:
- 0.689 mol * 44 g/mol = 30.3 g
Answer: the electric charge
Explanation:
Answer: 317 joules
Explanation:
The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
In this case,
Q = ?
Mass of aluminium = 50.32g
C = 0.90J/g°C
Φ = (Final temperature - Initial temperature)
= 16°C - 9°C = 7°C
Then, Q = MCΦ
Q = 50.32g x 0.90J/g°C x 7°C
Q = 317 joules
Thus, 317 joules of heat is gained.