Answer:
The volume of the concentrated nitric acid taken: <u>V₁ = 4.3 mL</u>
Explanation:
Given: Final pH of nitric acid (HNO₃) solution= 1.30
Volume of HNO₃ solution: V₂ = 600.0 mL
Concentration of stock HNO₃ solution: M₁ = 7.0 M
Volume of the stock HNO₃ solution: V₁ = ?
To find out <u>the concentration of HNO₃ solution (M₂),</u> we use the following equation: pH = - ㏒ [H⁺]
⇒ 1.3 = - ㏒ [H⁺]
⇒ ㏒ [H⁺] = - 1.3
⇒ [H⁺] = antilog (- 1.3)
⇒ [H⁺] = antilog (- 1.3) = 0.050 M = M₂
Now, to calculate<u> the volume of concentrated 7.0 M HNO₃ (V₁)</u> that should be added to prepare 600.0 mL of 0.050 M HNO₃ solution, we use the dilution equation: M₁ × V₁ = M₂ × V₂
⇒ V₁ = M₂ × V₂ ÷ M₁
⇒ V₁ = (0.050 M) × (600.0 mL) ÷ (7.0 M)
⇒ <u>V₁ = 4.3 mL (rounded to two significant digits)</u>
<u>Therefore, the volume of the concentrated nitric acid taken: V₁ = 4.3 mL</u>
Answer: A
Explanation: Boiling point increases with the rise of electrons.
Answer:
Option-4 (3:2) is the correct answer.
Explanation:
Following steps are taken to balance the given unbalanced chemical equation.
Step 1: Write the unbalanced chemical equation,
N₂ + H₂ → NH₃
Step 2: Balance Nitrogen Atoms;
There are 2 nitrogen atoms on left hand side and 1 nitrogen atoms on right hand site therefore, to balance them multiply NH₃ on right hand side by 2 i.e.
N₂ + H₂ → 2 NH₃
Step 3: Balance Hydrogen Atoms;
Now, there are 2 hydrogen atoms on left hand side and 6 hydrogen atom on right hand site therefore, to balance them multiply H₂ on left hand side by 3 i.e.
N₂ + 3 H₂ → 2 NH₃
Now, the equation is balanced.
Step 4: Finding out mole ratios:
From balanced chemical equation it can be concluded that 3 moles of H₂ are involved in producing 2 moles of NH₃ hence, the mole ratio of consumption of H₂ to production of NH₃ is 3:2.
Q must be supplied = 119523.3~J
<h3>Further explanation </h3>
The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released
Q in = Q out
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
So from the question :
Q to the system(Q supplied) = Q water + Q steel vessel(Q gained)