For the titration we use the equation,
M₁V₁ = M₂V₂
where M is molarity and V is volume. Substituting the known values,
(0.15 M)(43.2 mL) = (2)(M₂)(20.5 mL)
We multiply the right term by 2 because of the number of H+ in H2SO4. Calculating for M₂ will give us 0.158 M. Thus, the answer is approximately 0.16M.
Answer:
The least whole number coefficient for HNO₃ is 6
Explanation:
The chemical equation above is the reaction between calcium orthophosphate and nitric acid.
To balance a chemical equation, we have to consider law of conservation of matter which states that matter can neither be created nor destroyed.
What this law implies is that, whatever we have at the reactant side must be equal to whatever is obtainable at the product side.
The above equation is
Ca₃(PO₄)₂ + HNO₃ → Ca(NO₃)₂ + H₃PO₄
To balance the equation, we'll have to check the number of atoms at each side and possibly balance the equation with the number of moles.
The balanced equation is
Ca₃(PO₄)₂ + 6HNO₃ → 3Ca(NO₃)₂ + 2H₃PO₄
From the balanced equation above, we can see that the number of calcium (Ca), Phosphorus (P), Oxygen(O), Nitrogen(N) and hydrogen (H) are balanced at both sides of the equation.
The least number coefficient for HNO₃ is 6
Answer:
The molar mass of the unknown acid is 386.8 g/mol
Explanation:
Step 1: Data given
Mass of the weak acid = 1.168 grams
volume of NaOH = 28.75 mL = 0.02875 L
Molarity of NaOH = 0.105 M
Since we only know 1 equivalence point, we suppose the acid is monoprotic
Step 2: Calculate moles NaOH
Moles NaOH = molarity NaOH * volume NaOH
Moles NaOH = 0.105 M * 0.02875 L
Moles NaOH = 0.00302 moles
We need 0.00302 moles of weak acid to neutralize the NaOH
Step 3: Calculate molar mass of weak acid
Molar mass = mass / moles
Molar mass = 1.168 grams / 0.00302 moles
Molar mass = 386.8 g/mol
The molar mass of the unknown acid is 386.8 g/mol
With your mind. Boom. Lol I'm sorry I don't know I just need points
Answer:
there are approximately n ≈ 10²² moles
Explanation:
Since the radius of the earth is approximately R=6378 km= 6.378*10⁶ m , then the surface S of the earth would be
S= 4*π*R²
since the water covers 75% of the Earth's surface , the surface covered by water Sw is
Sw=0.75*S
the volume for a surface Sw and a depth D= 3 km = 3000 m ( approximating the volume through a rectangular shape) is
V=Sw*D
the mass of water under a volume V , assuming a density ρ= 1000 kg/m³ is
m=ρ*V
the number of moles n of water ( molecular weight M= 18 g/mole = 1.8*10⁻² kg/mole ) for a mass m is
n = m/M
then
n = m/M = ρ*V/M = ρ*Sw*D/M = 0.75*ρ*S*D/M = 3/4*ρ*4*π*R² *D/M = 3*π*ρ*R² *D/M
n=3*π*ρ*R² *D/M
replacing values
n=3*π*ρ*R² *D/M = 3*π*1000 kg/m³*(6.378*10⁶ m)² *3000 m /(1.8*10⁻² kg/mole) = 3*π*6.378*3/1.8 * 10²⁰ = 100.18 * 10²⁰ ≈ 10²² moles
n ≈ 10²² moles
