Answer:
0.185M sulfuric acid
Explanation:
Based on the reaction:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
<em>1 mole of sulfuric acid reacts with 2 moles of KOH</em>
Initial moles of H₂SO₄ and KOH are:
H₂SO₄: 0.750L ₓ (0.470mol / L) = <em>0.3525 moles of H₂SO₄</em>
KOH: 0.700L ₓ (0.240mol / L) = <em>0.168 moles of KOH</em>
The moles of sulfuric acis that react with KOH are:
0.168mol KOH ₓ (1 mole H₂SO₄ / 2 moles KOH) = 0.0840 moles of sulfuric acid.
Thus, moles that remain are:
0.3525moles - 0.0840 moles = <em>0.2685 moles of sulfuric acid remains</em>
As total volume is 0.700L + 0.750L = 1.450L, concentration is:
0.2685mol / 1.450L = <em>0.185M sulfuric acid</em>
Answer:
2578.99 years
Explanation:
Given that:
100 g of the wood is emitting 1120 β-particles per minute
Also,
1 g of the wood is emitting 11.20 β-particles per minute
Given, Decay rate = 15.3 % per minute per gram
So,
Concentration left can be calculated as:-
C left =
Where,
is the concentration at time t
is the initial concentration
Also, Half life of carbon-14 = 5730 years
Where, k is rate constant
So,
The rate constant, k = 0.000120968 year⁻¹
Time =?
Using integrated rate law for first order kinetics as:
So,
<u>t = 2578.99 years</u>
Answer:
1.40 M [OH⁻]
Explanation:
This compound dissociates into 3 ions, but since we are asked about [OH⁻], it's only 2. Therefore, multiply the molarity of the solution by the number of ions that [OH⁻] dissociates into:
2 × 0.70 M = 1.40 M
Hope this helps! Sorry that you got a link. Those are getting really annoying