All categories

3 years ago

If a solution containing

Chemistry

1 answer:

tankabanditka [31]3 years ago

6 0

Answer:

Hg(NO₃)₂(aq) + Na₂SO₄(aq) → 2NaNO₃(aq) + HgSO₄(s)

Moles of Hg(NO₃)₂ = 55.42 / 324.7 ==> 0.1707 moles

Moles of Na₂SO₄ = 16.642 / 142.04 ==> 0.1172 moles

Limiting reagent is Na₂SO₄ as it controls product formation

Moles of HgSO₄ formed = 0.1172 moles

= 0.1172 x 296.65

= 34.757g

Explanation:

You might be interested in

The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.380 g sample of liquid CHCl3 is placed in a closed,

Katyanochek1 [597]

Answer:

100 mm Hg

Explanation:

From the question we are told that

The vapor pressure of CHCl3, is $P = 100 \ mmHg = \frac{100}{760}= 0.13156 \ atm$

The temperature of CHCl3 is $T = 283 \ K$

The volume of the container is $V_c = 380mL = 380 *10^{-3}\ L$

The temperature of the container is $T_c = 283 \ K$

The mass of CHCl3 is m = 0.380 g

Generally the number of moles of CHCl3 present before evaporation started is mathematically represented as

$n = \frac{m }{M }$

Here M is the molar mass of CHCl3 with the value $M = 119.38 \ g/mol$

=> $n = \frac{ 0.380 }{119.38 }$

=> $n = 0.00318 \ mols$

Generally the number of moles of CHCl3 gas that evaporated is mathematically represented as

$n_g = \frac{PV}{RT}$

Here R is the gas constant with value $R = 0.08206 L \ atm /mol\cdot K$

$n_g = \frac{0.13156* 380 *10^{-3} }{0.08206 * 283}$

$n_g = 0.00215 \ mols$

Given that the number of moles of CHCl3 evaporated is less than the number of moles of CHCl3 initially present , then it mean s that not all the liquid evaporated

At equilibrium the temperature of CHCl3 will be equal to the pressure of air so the pressure at equilibrium is 100 mmHg

4 0

3 years ago

What is your streak and what fact is it?

kobusy [5.1K]

Answer:

unknown fact I can't message anyone

6 0

2 years ago

Read 2 more answers

Balance the equation and state the limiting reagent in the following reaction:

romanna [79]

Hgffvvcbchcgivigcigxgsgixigdfhcgcc fcfuc

8 0

3 years ago

Find the percent composition of each element in KMnO4

VARVARA [1.3K]

The way you want to find the percent composition would be by breaking down the problem like so:

K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999

Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO

Then you want to take each molar mass and then divide it 110.035 and multiply by 100

Ex. K = 39.098/ 110.035 and the multiply what you get by a 100

You do this for the other elements as well good luck!

6 0

3 years ago

Which property of a body cannot be changed if a force is applied to it?

prohojiy [21]

The answer is it’s mass

8 0

3 years ago

Read 2 more answers