Answer:
Explanation:
Calcium chloride is a soluble salt which dissociates into calcium and chloride ions when dissolved in water.
CaCl₂(aq) ----> Ca²⁺(aq) + 2Cl⁻(aq)
Similarly, sodium oxalate when dissolved in water dissociates into sodium and oxalate ions.
Na₂CO₄(aq) ----> 2Na⁺(aq) + C₂O₄²⁻(aq)
However, in a double displacement reaction where the two solutions of the salts are mixed, the insoluble salt calcium oxalate is precipitated. The net ionic equation for the reaction is shown below:
Ca²⁺(aq) + C₂O₄²⁻(aq) ----> CaC₂O₄(s)
The answer is C. The payoff of most risks is insignificant.
Answer:
Number of moles nitric acid in the cylinder is 400.539g/mol.
Explanation:
From the given,
Weight of empty gas cylinder
= 30.01 g/mol
Number of moles nitric acid =
=?
The mass of nitric acid in the cylinder = 

Number of moles of nitric acid =

Therefore, number of moles nitric acid in the cylinder is 400.539g/mol.
Answer:
296.1 day.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).
t is the time of the reaction (t = ??? day).
a is the initial concentration of Ir-192 (a = 560.0 dpm).
(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).
<em>∴ kt = lna/(a-x)</em>
(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).
(9.365 x 10⁻³ day⁻¹)(t) = 2.773.
<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>
Answer:
ΔU=-369.2 kJ/mol.
Explanation:
We start from the equation:
Δ(H)=ΔU+Δ(PV), which is an extension of the well known relation: H=U+PV.
If Δ(PV) were calculated by ideal gas law,
PV=nRT
Δ(PV)=RTΔn.
Where Δn is the change of moles due to the reaction; but, this reaction does not give a moles change (Four moles of HCl produced from 4 moles of reactants), so Δ(PV)=0.
So, for this case, ΔH=ΔU.
The enthalpy of reaction given is for one mole of reactant, so the enthalpy of reaction for the reaction of interest must be multiplied by two:

ΔU=-369.2 kJ/mol.