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:
Explanation:
Given parameters:
Initial temperature T₁ = 25.2°C = 25.2 + 273 = 298.2K
Initial pressure = P₁ = 0.6atm
Final temperature = 72.4°C = 72.4 + 273 = 345.4K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

where P and T are temperatures, 1 and 2 are initial and final temperatures.
Input the parameters and solve;
P₂ = 0.7atm
I will solve this question assuming the reaction equation look like this:
<span>MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O.
</span>
For every one molecule of MnO2 used, there will be one molecule of Cl2 formed. If the molecular mass of MnO2 is 87g/mol and molecular mass of Cl2 is <span> 73.0 g/mol, the mass of MnO2 needed would be:
Cl mass/Cl molecular mass * MnO2 molecular mass=
25g/ (73g/mol) * (87g/mol) * 1/1= 29.8 grams</span>
Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present
q2)
next we are asked to calculate the number of moles of water present
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present
sum of the products of the molar masses of the individual elements by the number of atoms
H - 1 g/mol and O - 16 g/mol
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol
molar mass of H₂O is 18 g/mol
therefore number of moles of water = 3.6 g / 18 g/mol = 0.2 mol
0.2 mol of water is present