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podryga [215]
3 years ago
10

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f

ill this container given a 49.4% CO2 recovery, assuming the other contents in the bag take up a negligible volume compared to the gas. The temperature of the room is 24.5 °C and the atmospheric pressure is 753.5 mmHg.
Chemistry
1 answer:
dsp733 years ago
5 0

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

(1 atm = 760 mmHg)

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65  K

Putting values in above equation, we get:

(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole

Percentage recovery of carbon dioxide gas =  49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

\frac{49.4}{100}\times  0.0438 mol=0.02164 mol

2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

\frac{2}{1}\times 0.02164 mol=0.04328 mol

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

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In which of the following cases does the reaction go to farthest to completion ?
Mashcka [7]

Answer:

k=10^{3}

Explanation:

k stand for equilibrium constants in terms of reaction

The higher the value of an equilibrium constant the faster the equilibrium reaction comes to completion.

Consider the example below:

R_{1}\ + R_{2}⇄P_{1}\ + P_{2}

where

The\ equilibrium\ constant\ K = \frac{P_{1}P_{2}}{R_{1}R_{2}}

For a faster reaction the numerator i.e. the right hand side of the equation have to be higher than the left hand side (the denominator). therefore the higher the numerator, the higher the value of the equilibrium constant and the faster the reaction get to completion thus option c is correct.

5 0
3 years ago
A balloon filled with 1.22 L of gas at 286 K is heated until the
TiliK225 [7]

Answer: 670K

Explanation:

Given that,

Original volume of gas V1 = 1.22 L

Original temperature T1 = 286 K

New volume V2 = 2.86 L

New temperature T2 = ?

Since volume and temperature are involved while pressure is constant, apply the formula for Charles law

V1/T1 = V2/T2

1.22 L/286 K = 2.86 L/ T2

Cross multiply

1.22 L x T2 = 286 K x 2.86 L

1.22T2 = 817.96

Divide both sides by 1.22

1.22T2/1.22 = 817.96/1.22

T2 = 670.459 K (Round to the nearest whole number as 670 K)

Thus, the temperature of the gas is 670 Kelvin

4 0
3 years ago
Write the equation for the reaction described: A solid metal oxide, Ag2O, and hydrogen are the products of the reaction between
Harlamova29_29 [7]

Answer:

2Ag + H2O -----> Ag2O + 2H

Explanation:

2Ag + H2O -----> Ag2O + 2H is the equation of the reaction between metal and steam. Silver reacts with water (steam) forming silver oxide and hydrogen gas. When the metals react with steam it produces the solid metal oxide and hydrogen gas. On the surface o metals, a protective layer of aluminium oxide is formed that keeps water away from the metal so we can say that silver oxide and hydrogen are formed from the reaction of silver metal and steam.

8 0
3 years ago
How many grams of mg are in 534g of mgo<br> ​
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Answer:

0.534mg..

....

...

..

...

.

....

7 0
2 years ago
Hund's rule states that electrons must spread out within a given subshell before they can pair
Temka [501]

Answer:

Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.

Explanation:

If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.

6 0
2 years ago
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