In molten lava like all the rest.
Hope it helped!!!
Answer:
We need 12.26 grams H2SO4
Explanation:
Step 1: Data given
Volume of a H2SO4 solution = 500 mL = 0.500 L
Concentration of the H2SO4 solution = 0.250 M
Molar mass of H2SO4 = 98.08 g/mol
Step 2: Calculate moles H2SO4
Moles H2SO4 = concentration * volume
Moles H2SO4 = 0.250 M * 0.500 L
Moles H2SO4 = 0.125 moles
Step 3: Calculate mass of H2SO4
Mass of H2SO4 = moles * molar mass
Mass of H2SO4 = 0.125 moles * 98.08 g/mol
Mass of H2SO4 = 12.26 grams
We need 12.26 grams H2SO4
PbSO₄ partially dissociates in water. the balanced equation is;
PbSO₄(s) ⇄ Pb²⁺(aq) + SO₄²⁻(aq)
Initial - -
Change -X +X +X
Equilibrium X X
Ksp = [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸ = X * X
1.6 x 10⁻⁸ = X²
X = 1.3 x 10⁻⁴ M
Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M.
[Pb²⁺] = 1.3 x 10⁻⁴ M.
= 1.3 x 10⁻⁴ mol / L x 207 g / mol
= 26.91 ppm
C = 12
O2 = 16*2= 32
CO2 = (12)+(16*2) = 44
32/44*100 = 72.73%
Answer:
6.48 L
Explanation:
From the question,
Applying
PV/T = P'V'/T'......................... Equation 1
P = initial pressure of the helium balloon, V = Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.
make V' the subject of the equation
V' = PVT'/P'T......................... Equation 2
Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm
Substitute these values into equation 2
V' = (4.5×1×253)/(0.6×293)
V' = 1138.5/175.8
V' = 6.48 L
