MgCO3+HCl->MgCl2+H2CO3

A glass container was initially charged with 2.00 moles of a gas sample at 3.75 atm and 21.7 °C. Some of the gas was released as

finlep [7]

Answer:

0.521 moles still present in the container.

Explanation:

It is possible to answer this question by using the general gas law, that is:

PV = nRT

Where P represents pressure of the gas, v its volume, n moles, R gas constant law and T absolute temperature (21.7°C + 273.15 = 294.85K)

Replacing with values of the initial conditions of the container, its volume is:

V = nRT / P

V = 2.00mol*0.082atmL/molK*294.85K / 3.75atm

V = 12.9L

When some gas is released, absolute temperature is 28.1°C + 273.15 = 301.25K, the pressure is 0.998atm and the volume of the container still constant. Again, using general gas law:

PV / RT = n

0.998atm*12.9L / 0.082atmL/molK*301.25K = n

0.521 moles = n

<h3>0.521 moles still present in the container.</h3>

8 0

2 years ago

The table below gives some characteristics of element Y. Characteristics of element Y Values Mass number 11 Number of neutrons i

Elden [556K]

I believe the answer is Na because it should be sodium which is Na.

6 0

2 years ago

Read 2 more answers

5. Gabi has plans with her friends to go to a concert on her birthday in 4 days. She is so excited that she wants to know how ma

drek231 [11]

Answer:

So she is very anxious because she has to wait 345600 seconds

Explanation:

60 second = 1 minute

60 minute = 1 hour

1 hour has 3600 seconds (60*60)

24 hour = 1 day

3600 second * 24 hours =

1 day has 86400 seconds so in four days

86400 * 4 = 345600

7 0

2 years ago

Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh

Naya [18.7K]

Answer:

1.82 g is the maximum mass of CuS.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $CuCl_2$ :

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of $CuCl_2$ :

$Moles=0.500 \times {38.0\times 10^{-3}}\ moles$

Moles of $CuCl_2$ = 0.019 moles

For $(NH_4)_2S$ :

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of $(NH_4)_2S$ :

$Moles=0.600 \times {42.0\times 10^{-3}}\ moles$

Moles of $(NH_4)_2S$ = 0.0252 moles

According to the given reaction:

$CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}$

1 mole of $CuCl_2$ reacts with 1 mole of $(NH_4)_2S$

So,

0.019 mole of $CuCl_2$ reacts with 0.019 mole of $(NH_4)_2S$

Moles of $(NH_4)_2S$ = 0.019 mole

Available moles of $(NH_4)_2S$ = 0.0252 mole

Limiting reagent is the one which is present in small amount. Thus, $CuCl_2$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of $CuCl_2$ gives 1 mole of $CuS$

0.019 mole of $CuCl_2$ gives 0.019 mole of $CuS$

Moles of $CuS$ formed = 0.019 moles

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

1.82 g is the maximum mass of CuS.

5 0

2 years ago

When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C

UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

$[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M$

Equilibrium constant of the reaction :

$K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}$

K = 4

$CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)$

Concentration at eq'm:

0.24 M 0.24 M 0.48 M 0.48 M

After addition of 0.34 moles per liter of $CO_2$ and $H_2$ are added.

(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

$K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}$

$4=\frac{(0.48+x)^2}{(0.24+0.34)^2}$

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0

2 years ago