A. We can calculate the initial concentrations of each by
the formula:
initial concentration ci = initial volume * initial
concentration / total mixture volume
where,
total mixture volume = 10 mL + 20 mL + 10 mL + 10 mL = 50
mL
ci (acetone) = 10 mL * 4.0 M / 50 mL = 0.8 M
ci (H+) = 20 mL * 1.0 M / 50 mL = 0.4 M (note: there is only 1 H+ per
1 HCl)
ci (I2) = 10 mL * 0.0050 M / 50 mL = 0.001 M
B. The rate of reaction is determined to be complete when
all of I2 is consumed. This is signified by complete disappearance of I2 color
in the solution. The rate therefore is:
rate of reaction = 0.001 M / 120 seconds
rate of reaction = 8.33 x 10^-6 M / s
Copper oxide is the only product, and it contains copper and oxygen.
one of the reactants is copper, so the other reactant must be oxygen.
The copper metal must have combined with something in the air.
Answer:
yes, soil is weathered from the bedrock below the subsoil. Soil contains whatever the bedrock below it consist of whatever the bedrock below it is made of.
Explanation:
Hope this helps!
M(Ag)=12.5 g
Nₐ=6.022*10²³ mol⁻¹
n(Ag)=m(Ag)/M(Ag)
N=n(Ag)*Nₐ
N=Nₐm(Ag)/M(Ag)
N=6.022*10²³mol⁻¹*12.5g/107.868g/mol=6.97*10²²
A. 6.97 × 10²² atoms
Answer:
2.30 liters.
Explanation:
- The balanced equation of the reaction is:
<em>Na₂O₂ + CO₂ → Na₂CO₃ + 1/2O₂,</em>
- It is clear that 1.0 mole of Na₂O₂ reacts with 1.0 mole of CO₂ to produce 1.0 mole of Na₂CO₃ and 0.5 mole of O₂.
- The no. of moles of CO₂ in (4.60 L) reacted can be calculated from the relation: <em>PV = nRT</em>.
P is the pressure of the gas (P = 1.0 atm at STP),
V is the volume of the gas (V = 4.60 L),
R is the general gas constant (R = 0.082 L.atm/mol.K),
T is the temperature of the gas (T = 273.0 K at STP).
∴ n = PV/RT = (1.0 atm)(4.6 L) / (0.082 L.atm/mol.K)(273.0 K) = 0.205 mol.
<u><em>Using cross multiplication:</em></u>
1.0 mole of CO₂ produces → 0.5 mole of O₂, from the stichiometry.
0.205 mole of CO₂ produces → ??? mole of O₂.
- The no. of moles of O₂ produced from 4.60 L of CO₂ = (0.5 mole)(0.205 mole) / (1.0 mole) = 0.103 mole.
- ∴ The volume of O₂ produced from 4.60 L of CO₂ = nRT/P = (0.103 mol)(0.082 L.atm/mol.K)(273.0 K) / (1.0 atm) = 2.30 liters.
