Answer:
0.3229 M HBr(aq)
0.08436M H₂SO₄(aq)
Explanation:
<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>
<em />
Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).
NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)
When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.
<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>
<em />
Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).
2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)
When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.
Answer:
Conduct electricity when they are molten, while covalent compounds usually do not conduct electricity when they are molten.
% yield = 80.719
<h3>Further explanation</h3>
Given
22.0 g of Mgl₂
25.0 g of Mg
25.0 g of l₂
Required
The percent yield
Solution
Reaction
Mg + I₂⇒ MgI₂
mol Mg = 25 g : 24.305 g/mol = 1.029
mol I₂ = 25 g : 253.809 g/mol = 0.098
Limiting reactant = I₂
Excess reactant = Mg
mol MgI₂ based on I₂, so mol MgI₂ = 0.098
Mass MgI₂ (theoretical):
= mol x MW
= 0.098 x 278.114
= 27.255 g
% yield = (actual/theoretical) x 100%
% yield = (22 / 27.255) x 100%
% yield = 80.719
The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.
The volume of water is 
.
Since, 1 gal= 3785.41 mL
Thus, 
Density of water is 1 g/mL thus, mass of water will be 
.
Since, 1 grams of chlorine →
grams of water.
1 g of water →
g of chlorine and,
of water →86.6 g of chlorine
Since, the solution is 9% chlorine by mass, the volume of solution will be:
Thus, volume of chlorine solution is 9.62\times 10^{2} mL.
Answer:
n NaHCO3 = 9.6 E-3 mol
Explanation:
balanced reaction:
- 2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)
- assuming a concentration of H2SO4 6M....normally worked in the lab
⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4
according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)
⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )
⇒ ,mol NaHCO3 = 9.6 E-3 mol
So 9.6 E-3 mol NaHCO3, are the minimun moles necessary to neutralize the acid.
