At pH 3.70, H2SO4 can be regarded as being fully dissociated. The dissociation reaction is
H₂SO₄(l) --> 2H⁺(aq) + SO₄²⁻(aq)
Since pH = -log₁₀[H⁺]
[H⁺] = 
= 
[H⁺] = 2 x 10⁻⁴ M
Since there are 2 moles of H⁺ for every mole of H₂SO₄
[H₂SO₄] = 0.5 * [H⁺]
=0.5 * 2 x 10⁻⁴ M
[H₂SO₄] = 1 x 10⁻⁴ M
Convert the amount of rain and the area into SI units
Rain = 1.00 in * 0.0254 m/in
Rain = 0.0254 m
Area = 1600 miles² * 2589988.11 m²/miles² = 4.661 x 10⁹ m²
Calculate the volume of water
Volume = Rain * Area
=0.0254 m * 4.661 x 10⁹
Volume = 1.184 x 10⁸ m³
Calculate the number of kilo moles of H₂SO₄
kilo moles H₂SO₄ = [H₂SO₄] * Volume
=1.0 x 10⁻⁴ M * 1.184 x 10⁸ m³
kilo moles H₂SO₄ = 11,839 k mole
Calculate the mass of H₂SO₄
mass H₂SO₄ = kilo moles H₂SO₄ * MW H₂SO₄
= 11839 k mol * 98.07848 kg/kmol
mass H₂SO₄ = 1.16 x 10⁶ kg H₂SO₄
Lack of cell walls no chloroplasts...hoped that helped
Answer:
c = 4
Explanation:
In general, for the reaction
a A + b B ⇒ c C + d D
the rate is given by:
rate = - 1/a ΔA/Δt = - 1/b ΔB/Δt = + 1/c ΔC/Δt = + 1/d ΔD/Δt
this is done so as to express the rate in a standarized way which is the same to all the reactants and products irrespective of their stoichiometric coefficients.
For this question in particular we know the coefficient of A and need to determine the coefficient c.
- 1/2 ΔA/Δt = + 1/c ΔC/Δt
- 1/2 (-0.0080 ) = + 1/c ( 0.0160 mol L⁻¹s⁻¹ )
0.0040 mol L⁻¹s⁻¹ c = 0.0160 mol L⁻¹s⁻¹
∴ c = 0.0160 / 0.0040 = 4
Answer:
1.2 × 10^22 atoms.
Explanation:
Firstly, cations refers to the positively charged atom in the ionic compound, which is Na+.
Given the mass of NaCl as 1.17g, the number of moles of NaCl can be calculated this:
Molar mass of NaCl = 23 + 35.5
= 58.5g/mol
Mole = mass/molar mass
Mole = 1.17/58.5
Mole = 0.02moles
Using Avagadro's number, 6.022 × 10^23 atoms of Na+ are in 1 mole of NaCl.
In 0.02 moles of NaCl, there are 0.02 × 6.022 × 10^23 of Na+
0.1204 × 10^23 atoms
1.2 × 10^22 atoms of Na+ (cation)