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brilliants [131]
2 years ago
6

1. Consider the following: h2so4 2 naoh → na2so4 2 h2o if a 24. 0 ml sample of h2so4 (aq) was neutralized by 20. 0 ml of 0. 125

m naoh, calculate the molarity of the acid sample
Chemistry
1 answer:
jasenka [17]2 years ago
8 0

The molarity of the acid sample H₂SO₄ is 0.052M .

<h3>What is Molarity ?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution.

Molarity is defined as the moles of a solute per liters of a solution.

Molarity is also known as the molar concentration of a solution

Now to determine the molarity of the acid sample

V( H₂SO₄) = 24.0 mL in liters = 24.0 / 1000 = 0.024 L

M(H₂SO₄) = ?

V(NaOH) = 20.0 mL = 20.0 / 1000 = 0.02 L

M(NaOH) = 0.125 M

Number of moles NaOH :

n = M x V

n = 0.125 x  0.02

n = 0.0025 moles of NaOH

H₂SO₄(aq) + 2 NaOH(aq) = Na₂SO₄(aq) + 2 H₂O(l)

1 mole H₂SO₄ ---------- 2 mole NaOH

? mole H₂SO₄ ---------- 0.0025 moles NaOH

moles = 0.0025 * 1 / 2

= 0.00125 moles of H₂SO₄

M(H₂SO₄) = n / V

M = 0.00125 /  0.024

= 0.052 M

Therefore the molarity of the acid sample H₂SO₄ is 0.052M .

To know more about molarity

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4 0
2 years ago
Calculate the total energy, in kilojoules, that is needed to turn a 46 g block
neonofarm [45]

Answer: The total energy, in kilojoules, that is needed to turn a 46 g block of ice at -25 degrees C into water vapor at 100 degrees C is 11.787 kJ.

Explanation:

Given: Mass = 46 g

Initial temperature = -25^{o}C

Final temperature = 100^{o}C

Specific heat capacity of ice = 2.05 J/g^{o}C

Formula used to calculate the energy is as follows.

q = m \times C \times (T_{2} - T_{1})

where,

q = heat energy

m = mass

C = specific heat capacity

T_{1} = initial temperature

T_{2} = final temperature

Substitute the values into above formula as follows.

q = m \times C \times (T_{2} - T_{1})\\= 46 g \times 2.05 J/g^{o}C \times (100 - (-25))^{o}C\\= 11787.5 J (1 J = 0.001 kJ)\\= 11.787 kJ

Thus, we can conclude that the total energy, in kilojoules, that is needed to turn a 46 g block of ice at -25 degrees C into water vapor at 100 degrees C is 11.787 kJ.

7 0
3 years ago
NEED HELP ASAP NOT DIFFICULT
Burka [1]
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
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