molar concentration of AgNO₃ solution = 0.118 mole/L
Explanation:
Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.
number of moles = mass / molecular weight
number of moles of AgNO₃ = 10 / 170 = 0.0588
molar concentration = number of moles / volume (L)
molar concentration of AgNO₃ solution = 0.0588 / 0.5
molar concentration of AgNO₃ solution = 0.118 mole/L
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molar concentration
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Answer:
5 moles of oxygen are required.
Explanation:
Given data:
Moles of O₂ required = ?
Moles of H₂ present = 10 mol
Solution:
Chemical equation:
O₂ + 2H₂ → 2H₂O
Now we will compare the moles of oxygen and hydrogen.
H₂ : O₂
2 : 1
10 : 1/2×10 = 5 mol
5 moles of oxygen are required.
Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
Putting the values we get :
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
A solution of K2SO4 and KCl is added to a solution of Ba(NO3)2. BaS0₄ (s) will precipitate out of this combined solution.
Molecular equation
K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2 KNO3(aq)?
This equation represents a double displacement (replacement) reaction, also called a metathesis reaction, in which the reactant ions exchange places to form new products. The general equation is:
A-B + C-D → A-D + C-B;
where A and C are cations, and B and D are anions.
Complete ionic equation: Includes all ions and the precipitate.
2K^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2[NO3]^-(aq) → 2K^+(aq) + 2[NO3]^- + BaSO4(s)
In an aqueous solution, precipitation is the process of transforming a dissolved substance into an insoluble solid from a super-saturated solution.
The solid formed is called the precipitate. In case of an inorganic chemical reaction leading to precipitation, the chemical reagent causing the solid to form is called the precipitant.
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Answer:
'See Explanation
Explanation:
Determine the [OH−] , pH, and pOH of a solution with a [H+] of 9.5×10−13 M at 25 °C.
Given [H⁺] = 9.5 x 10⁻¹³M => [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ => [OH⁻] = 1.0 x 10⁻¹⁴/9.5 x 10⁻¹³ = 0.0105M
pH = -log[H⁺] = -log(9.5 x 10⁻¹³) = - (-1202) = 12.02.
pOH = -log[OH⁻] = -log(0.0105) = -(-1.98) = 1.98
Now you use the same sequence in the remaining problems.