Answer:
Explanation:
Ksp(BaSO4)=1.07×10−10
BaSO₄ → Ba²⁺ + SO₄²⁻
1.07×10⁻¹⁰ = ( Ba²⁺) × ( SO₄²⁻)
but Ba²⁺ = 1.3×10⁻² M
1.07×10⁻¹⁰ = 1.3×10⁻² M × ( SO₄²⁻)
( SO₄²⁻) = 1.07×10⁻¹⁰ / 1.3×10⁻² = 0.823 × 10⁻⁸ M
while Ksp(CaSO4)=7.10×10−5
CaSO₄ → Ca²⁺ + SO₄²⁻
7.10×10⁻⁵ = 2.0×10⁻² × ( SO₄²⁻)
( SO₄²⁻) = 7.10×10⁻⁵ / 2.0×10⁻² = 3.55 × 10⁻³ M
comparing the concentration of sulfate ions, Ba²⁺ cation will precipitate first because the Ba²⁺ requires 0.823 × 10⁻⁸ M sodium sulfate which less compared the about needed by CaSO₄
Answer:
THE NEW VOLUME OF THE GAS IS 406 mL WHEN THE TEMPERATURE CHANGES FROM 765 K TO 315 K.
Explanation:
When the temperature changes from 765 K to 315K, the volume has changed from 986 mL to?
V1 = 986 mL = 0.986 L
T1 = 765 K
T2 = 315 K
V2 = unknown
Using Charles' equation of gas laws;
V1 / T1 = V2 / T2
Making V2 the subject of the formula:
V2 = V1 T2 / T1
V2 = 0.986 * 315 / 765
V2 = 0.406 L
V2 = 406 mL
So therefore, the volume of a gas changes from 986 mL to 406 mL as a result of a change in temperature from 765 K to 315 K.
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹
Answer:
The unknown temperature is 304.7K
Explanation:
V1 = 100mL = 100*10^-3L
P1 = 99.10kPa = 99.10*10³Pa
V2 = 74.2mL = 74.2*10^-3L
P2 = 133.7kPa = 133.7*10³Pa
T2 = 305K
T1 = ?
From combined gas equation,
(P1 * V1) / T1 = (P2 * V2) / T2
Solving for T1,
T1 = (P1 * V1 * T2) / (P2 * V2)
T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)
T1 = 3022550 / 9920.54
T1 = 304.67K
T1 = 304.7K
Stars born larger than 8 solar masses usually retain enough mass to undergo core collapse, with the resulting shock wave producing a Type Ib supernova (spectra without Hydrogen or Silicon lines, with Helium lines), a Type Ic supernova (without Hydrogen or Helium or Silicon lines) or a Type II supernova
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