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Elden [556K]
3 years ago
11

A solution contains 0.2 moles of hydrochloric acid. The volume of the solution is 100cm3. What is its concentration in mol/dm³?

Chemistry
1 answer:
Umnica [9.8K]3 years ago
4 0
Answer= 2 mol/dm^3

Explanation:-
Firstly, we convert the volume, which is in cm^3 to dm^3. To do that divide number in cm^3 by 1000:
100/1000= 0.1 dm^3

Formula for concentration is:
Concentration= No. of moles/ Volume

So,
Concentration= 0.2/0.1

Answer= 2 mol/dm^3

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16. What is tincture of iodine?​
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Answer:

Tincture of iodine, iodine tincture, or weak iodine solution is an antiseptic. It is usually 2–7% elemental iodine, along with potassium iodide or sodium iodide, dissolved in a mixture of ethanol and water. Tincture solutions are characterized by the presence of alcohol.

Explanation:

7 0
3 years ago
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
liraira [26]

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

3 0
4 years ago
A solution is prepared by dissolving 0.23 mol of benzoic acid and 0.27 mol of sodium benzoate in water sufficient to yield 1.00
Monica [59]

Answer:

The pH does not decrease drastically because the NaOH reacts with the <u>D) Benzoic acid</u> present in the buffer solution.

Explanation:

The hydroxide ions will react with acidic part of the solutions, it means the benzoic acid, so it will form the conjugate base, the benzoate ion.

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The half-life of na-24 is 15 hours . when there are 1000 atoms of na-24 in a sample , a scientist starts a stopwatch . the scien
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125 Each half life it divides by 2 the amount
1000/2=500
500/2=250
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3 years ago
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