Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
Answer:
2.75 mol
Explanation:
Given data:
Mass of Nitrogen = 38.5 g
Moles of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 38.5 g/ 28 g/mol
Number of moles = 1.375 mol
Now we will compare the moles of ammonia and nitrogen from balance chemical equation.
N₂ : NH₃
1 : 2
1.375 : 2×1.375 = 2.75 mol
Thus 2.75 moles of ammonia are produced from 38.5 g of nitrogen.
1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon.
Hence, in this case the mass of carbon in 8.46 g of CO2:
(12/44) × 8.46 = 2.3073 g
1 mole of water contains 18 g, out of which 2 g is hydrogen;
Therefore, 2.6 g of water contains;
(2/18) × 2.6 = 0.2889 g of hydrogen.
Therefore, with the amount of carbon and hydrogen from the hydrocarbon we can calculate the empirical formula.
We first calculate the number of moles of each,
Carbon = 2.3073/12 = 0.1923 moles
Hydrogen = 0.2889/1 = 0.2889 moles
Then, we calculate the ratio of Carbon to hydrogen by dividing with the smallest number value;
Carbon : Hydrogen
0.1923/0.1923 : 0.2889/0.1923
1 : 1.5
(1 : 1.5) 2
= 2 : 3
Hence, the empirical formula of the hydrocarbon is C2H3
Answer:
40% of the energy release by the peanut is 3500 calories
Explanation:
One calorie is defined as the amount of energy required to increase the temperature of one gram of water for one degree Celsius (or one Kelvin)
Equation for energy gain by water is
Q = mcΔT
where, m is the mass of the object
c is the specific heat capacity
ΔT is the change in temperature
c = 1.0 cal/g?°C.
m = 50 g
ΔT = 50°C - 22°C
= 28°C
Q = (50)× (1)× (28)
= 1400calories
The peanut contain 1400calories of energy .
amount that 40% of energy is released to water ,
so,
Q = 1400 calories / 0.4
= 3500 calories
Therefore, 40% of the energy release by the peanut is 3500 calories
