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VMariaS [17]
2 years ago
9

How many moles of na+ are present in 343 ml of a 1.27 m solution of na2so4?

Chemistry
1 answer:
cricket20 [7]2 years ago
4 0
Sodium sulfate, Na₂SO₄, is a strong electrolyte. It means that it dissociates into 2 Na⁺ ions and 1 SO₄⁻ ion when dissolved in water. So, for every 1 mole of Na₂SO₄, there is an equivalent amount of 2 mol Na⁺ ions. The solution is as follows:

Moles of Na⁺ = (1.27 mol Na₂SO₄/L)*(1 L/1000 mL)*(343 mL)*(2 mol Na⁺/1 mol Na₂SO₄)
<em>Moles of Na⁺ = 0.87 mol</em>
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0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A
elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = \frac{0.9975 g}{56 g/mol}=0.01781 mol

The enthlapy of solution :

\frac{Q'}{moles}

=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol

The enthlapy of solution is -55.23 kJ/mol.

8 0
2 years ago
Carbon tetrachloride contains one carbon and four chlorine atoms. for this compound 11.818 g of chlorine combine with 1.000 g of
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What is the name of the compound CuCO³?
Tems11 [23]
Basic copper carbonate
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The infections that people get while they are in hospitals are commonly resistant to multiple drugs. What is a reason for the in
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The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW =
Varvara68 [4.7K]

Answer:

9.744g of monosodium succinate.

4.925g of disodium succinate.

Explanation:

To find pH of the buffer produced by the mixture of monosodium succinate-Disodium succinate is obtained from H-H equation:

pH = pKa + log ([Na₂Suc] / [NaHSuc])

As you want a pH of 5.28 and pKa is 5.64:

5.28 = 5.64 + log ([Na₂Suc] / [NaHSuc])

-0.36 = log ([Na₂Suc] / [NaHSuc])

0.4365 = ([Na₂Suc] / [NaHSuc]) <em>(1)</em>

<em />

As total concentration of the buffer is 100mM = 0.100M:

0.100M = [Na₂Suc] + [NaHSuc] <em>(2)</em>

Replacing (2) in (1):

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 = (0.100M - [NaHSuc] / [NaHSuc])

0.4365 [NaHSuc] = 0.100M - [NaHSuc]

1.4365 [NaHSuc] = 0.100M

[NaHSuc] = 0.0696M

And:

[Na₂Suc] = 0.0304M

As volume of the buffer is 1L:

[NaHSuc] = 0.0696 moles

[Na₂Suc] = 0.0304 moles

Using molar mass of both substances:

Mass of monosodium succinate:

0.0696moles * (140g / 1mol) =<em> 9.744g of monosodium succinate.</em>

Mass of disodium succinate:

0.0304moles * (162g / 1mol) =<em> 4.925g of disodium succinate.</em>

<em></em>

5 0
2 years ago
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