Answer:
136.36 mL
Explanation:
Here we have to use the dilution formula
From C1V1= C2V2
Where;
C1= initial concentration of the solution= 12.0 M
C2= final concentration of the solution= 2.20 M
V1 = initial volume of the solution= 25.0 ml
V2= final volume of the solution= ?????
Then recall;
C1V1=C2V2
V2 = C1V1/C2
Substituting values from the parameters given;
V2= 12.0 × 25.0 / 2.20
V2= 136.36 mL
Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
A weak acid has a low concentration of H+ Ions and a dilute acid is a solution where acid is dissolved in a more volume of water than that of acid.
Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole
Moles = 0.246
Volume = 280 mL = 0.280 L
Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g
Moles = 1.98
Volume = 2.6 L
Molarity = 0.762 M
Molarity = 0.76 M
