Organic Chemistry- Substances only found in living organisms.
Physical Chemistry- Structure of physical compounds and the amount of energy they have.
Inorganic Chemistry- Study of carbon containing compounds.
Answer:
0.0303 Liters
Explanation:
Given:
Mass of the potassium hydrogen phosphate = 0.2352
Molarity of the HNO₃ Solution = 0.08892 M
Now,
From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃
The number of moles of 0.2352 g of potassium hydrogen phosphate
= Mass / Molar mass
also,
Molar mass of potassium hydrogen phosphate
= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol
Number of moles = 0.2352 / 174.15 = 0.00135 moles
thus,
The number of moles of HNO₃ required for 0.00135 moles
= 2 × 0.00135 mol of HNO₃
= 0.0027 mol of HNO₃
Now,
Molarity = Number of Moles / Volume
thus,
for 0.0027 mol of HNO₃, we have
0.08892 = 0.0027 / Volume
or
Volume = 0.0303 Liters
Answer:
<h3>The answer is 1.84 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>1.84 g/mL</h3>
Hope this helps you
Explanation:
The balanced equation of the reaction is given as;
Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)
1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?
From the reaction;
2 mol of HBr produces 1 mol of MgBr2
Converting to masses using;
Mass = Number of moles * Molar mass
Molar mass of HBr = 80.91 g/mol
Molar mass of MgBr2 = 184.113 g/mol
This means;
(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2
18.3g would produce x
161.82 = 184.113
18.3 = x
x = (184.113 * 18.3 ) / 161.82 = 20.8 g
2. How many moles of H2O will be produced from 18.3 grams of HBr?
Converting the mass to mol;
Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol
From the reaction;
2 mol of HBr produces 2 mol of H2O
0.226 mol would produce x
2 =2
0.226 = x
x = 0.226 * 2 / 2 = 0.226 mol
3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?
From the reaction;
2 mol of HBr reacts with 1 mol of Mg(OH)2
18.3g of HBr = 0.226 mol
2 = 1
0.226 = x
x = 0.226 * 1 /2
x = 0.113 mol
