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3 years ago

Calculate the volume of 1 M NaOH needed to prepare 10.0 ml of a 0.10 M solution.

1 answer:

7 0

Answer is: volume of 1 M NaOH is 1 ml.
c₁(NaOH) = 1 M.
V₂(NaOH) = 10 ml.
c₂(NaOH) = 0,1 M.
V₁(NaOH) = ?
c₁ - original concentration of the solution, before it gets diluted.
c₂ - final concentration of the solution, after dilution.
V₁ - volume to be diluted.
V₂ - final volume after dilution.
c₁ · V₁ = c₂ · V₂.
V₁(NaOH) = c₂ · V₂ ÷ c₁.
V₁(NaOH) = 0,1 M · 10 ml ÷ 1 M.
V₁(NaOH) = 1 ml.

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The answer is: the mass of oxygen is 16.95 grams.

The overall balanced photosynthesis reaction:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.

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n(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆).

n(C₆H₁₂O₆) = 15.9 g ÷ 180.18 g/mol.

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What is the pH of a solution whose hydronium ion [H20+] (or proton [H+1)

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Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

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Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

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