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maxonik [38]
3 years ago
5

What volume (L) of 0.250 M HNO3 is required to neutralize a solution prepared by dissolving 17.5 g of NaOH in 350 mL of water ?

Chemistry
1 answer:
IRINA_888 [86]3 years ago
5 0
<h3>Answer:</h3>

1.75 L HNO₃

<h3>Explanation:</h3>

We are given;

Molarity of HNO₃ as 0.250 M

Mass of NaOH as 17.5 g

Volume of NaOH = 350 mL

We are required to calculate the volume of 0.250 M

We are going to first write the balanced reaction:

NaOH(aq) + HNO₃(aq) → NaNO₃(aq) + H₂O(l)

Then, we calculate the number of moles of NaOH

Moles = Mass ÷ Molar mass

Molar mass of NaOH = 39.997 g/mol

          = 17.5 g ÷ 39.997 g/mol

          = 0.4375 moles

We can now calculate the number of moles of HNO₃ using the mole ratio from the equation;

Mole ratio of NaOH to HNO₃ is 1 : 1

Therefore, if moles of NaOH are 0.4375 moles then;

Moles of HNO₃ will also be 0.4375 moles

We can now calculate the volume of HNO₃

Morality = Number of moles ÷ Volume

Thus;

Volume = Number of moles ÷ Molarity

             = 0.4375 moles ÷ 0.250 M

             = 1.75 L

Therefore, the volume of HNO₃ is 1.75 L

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Which of the following is unchanged at the end of the CNO cycle?
Musya8 [376]

Answer:

The correct option is: A. carbon-12

Explanation:

The CNO cycle, the abbreviation for the carbon-nitrogen-oxygen cycle, is a catalytic cycle by which the stars produce helium from elemental hydrogen, via a series of nuclear fusion reactions.

This cycle involves the fusion of four protons with carbon (_{6}^{12}\textrm{C}), nitrogen isotope (_{7}^{13}\textrm{N}), and oxygen isotope (_{8}^{15}\textrm{O}), to give an alpha particle and two electron neutrinos and positrons.

The reaction involves the regeneration of carbon (_{6}^{12}\textrm{C}) nucleus in the last step.

_{6}^{12}\textrm{C} → _{7}^{13}\textrm{N} → _{6}^{13}\textrm{C} → _{7}^{14}\textrm{N} → _{8}^{15}\textrm{O} → _{7}^{15}\textrm{N} → _{6}^{12}\textrm{C}

3 0
3 years ago
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO
zavuch27 [327]

Answer:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons  4NOBr ( g )

The suitable equilibrium constant turns out:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Or in terms of the initial equilibrium constant:

K_2=K_1^2

Since the second reaction is a doubled version of the first one.

Best regards.

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3 years ago
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