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3 years ago

If aqueous solutions of sodium chloride and mercury(I) nitrate are mixed, which insoluble precipitate is formed?

Chemistry

1 answer:

NeX [460]3 years ago

7 0

Answer:

Hg2^2+(aq) + 2Cl^-(aq) —> Hg2Cl2(s)

Explanation:

The balanced equation for the reaction is given below:

2NaCl(aq) + Hg2(NO3)2(aq) —> 2NaNO3(aq) + Hg2Cl2(s)

Considering the states of each compound in the reaction, we can see that Hg2Cl2 is in solid form meaning it will precipitate out of the solution

In solution the following occurs:

NaCl —> Na+(aq) + Cl-(aq)

Hg2(NO3)2 —> Hg2^2+(aq) + 2NO3^-(aq)

Combining the two equation together, a balanced double displacement reaction occurs as shown below:

2Na+(aq) + 2Cl-(aq) + Hg2^2+(aq) + 2NO3^-(aq) —> 2Na+2NO3^-(aq) + Hg2^2+2Cl-(s)

From the above we can thus right the insoluble precipitate as

Hg2^2+(aq) + 2Cl^-(aq) —> Hg2Cl2(s)

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Is it possible to convert between moles and mass?

lubasha [3.4K]

Answer:

Yes, if you're talking about molar mass or grams

Explanation:

Hope this helps!

5 0

2 years ago

Read 2 more answers

PLSSS HELP. And there’s also at the bottom of the picture another option if u didn’t see it

WARRIOR [948]

Answer:

B. Gradualism

Explanation:

8 0

2 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

7 0

2 years ago

Iron sulfite has the formula?

Mademuasel [1]

Its FeSO3 or iron(iii)sulfite = Fe2(SO3)3

7 0

3 years ago

What mass of calcium carbonate is produced from 5.11<br> mol of sodium carbonate?

a_sh-v [17]

Answer:

There is 541.6 grams of calcium carbonate produced from 5.11 moles

Explanation:

Step 1: Data given

Number of moles of sodium carbonate ( Na2CO3) = 5.11 moles

Molar mass of sodium Carbonate:

⇒ Molar mass of Sodium = 22.99 g/mole

⇒ Molar mass of Carbon = 12.01 g/mole

⇒ Molar mass of Oxygen = 16 g/mole

Molar mass of Na2CO3 = 2* 22.99 + 12.01 + 3*16 = 105.99 g/mole

Step 2: Calculate mass of 5.11 moles of sodium carbonate

mass = number of moles of sodium carbonate * Molar mass of sodium carbonate

mass = 5.11 moles * 105.99 g/moles = 541.6 grams ≈ 542 grams

There is 541.6 grams of calcium carbonate produced from 5.11 moles Na2CO3

3 0

3 years ago