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olga_2 [115]
3 years ago
10

A self-contained underwater breathing apparatus (SCUBA) uses canisters containing potassium superoxide. The superoxide consumes

the CO2 exhaled by a person and replaces it with oxygen. 4 KO2(s) + 2 CO2(g) n 2 K2CO3(s) + 3 O2(g) What mass of KO2, in grams, is required to react with 8.90 L of CO2 at 22.0 °C and 767 mm Hg
Chemistry
1 answer:
Mrac [35]3 years ago
8 0

Answer:

52.0004 grams of mass of potassium superoxide  is required

Explanation:

Let moles carbon dioxide gas be n at 22.0 °C and 767 mm Hg occupying 8.90 L of volume.

Pressure of the gas,P = 767 mm Hg = 0.9971 atm

Temperature of the gas,T = 22.0 °C = 295.15 K

Using an ideal gas equation to calculate the number of moles.

PV=nRT

n=\frac{0.9971 atm\times 8.90 L}{0.0821 atm L/mol K\times 295.15 K}

n = 0.3662 mol

4KO_2(s)+2CO_2(g)\rightarrow 2K_2CO_3(s)+3O_2(g)

According to reaction, 2 moles of carbon-dioxide reacts with 4 moles of potassium superoxide.

Then 0.3662 mol of  of carbon-dioxide will react with:

\frac{4}{2}\times 0.3662 mol=0.7324 mol of potassium superoxide.

Mass of 0.7324 mol potassium superoxide:

0.7324 mol × 71 g/mol = 52.0004 g

52.0004 grams of mass of potassium superoxide is required.

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vaieri [72.5K]
Nitrogen dioxide is NO2 (where the number 2 is a subscript that indicates the number of atoms in one molecule).

Use Avogadro's number: 1 mol of molecules = 6.022 * 10^ 23 molecules.

So, in 5.6 mol of NO2 there are 5,6 mol * 6.022 * 10^23 molecules / mol = 33.72 * 10^23 molecules = 3.372 * 10^24 molecules of NO2.

There are the same number of N atoms (because there is 1 atom of N in 1 molecule) and there are the double of O atoms (becasue there are 2 atoms of O in 1 molecule) => 3.372 * 10 ^24 atomsof N and 6.744 atoms of O.

Answer: 3.372 * 10^24 molecules of NO2, 3.372 * 10^24 atoms of N and 6.744 atoms of O.
5 0
3 years ago
How many grams of water will be produced if you start with 4.0 grams of hydrogen and an excess of oxygen given the following bal
Ivahew [28]

Answer:

B.) 36.0 grams

Explanation:

To find the mass of water, you need to (1) convert grams H₂ to moles (using the molar mass), then (2) convert moles H₂ to moles H₂O (using mole-to-mole ratio from reaction coefficients), and then (3) convert moles H₂O to grams (using the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units.

Molar Mass (H₂): 2(1.008 g/mol)

Molar Mass (H₂): 2.016 g/mol

Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol

Molar Mass (H₂O): 18.014 g/mol

2 H₂ + O₂ -----> 2 H₂O
^                        ^

4.0 g H₂            1 mole             2 moles H₂O           18.014 g
---------------  x  ----------------  x  ------------------------  x  ---------------  = 36 g H₂O
                          2.016 g             2 moles H₂              1 mole

3 0
1 year ago
What happens when sugar is heated? Give a balanced equation
mario62 [17]

Answer:

when the sugar is heated it turn out in caramelize

8 0
2 years ago
The density of water at 4 degrees Celsius is 1.00 g/cm cubed at unknown object has density of 7.9 g/cm cubed would you expect it
NNADVOKAT [17]

Answer:

The object will be sank

Explanation:

In this case the object is more dense than water.

Density is the relationship between a certain amount of mass of matter and the volume that is being occupied by it.

The object occupies more volume, so it occupies more mass.

As the mass from the object is higher, the object will be sank because the weight is higher than the weight from the liquid.

If the object has a lower density than the water, it will float on it.

7 0
3 years ago
C-12 and C-13 are naturally-occurring isotopes of the element carbon. C-12 occurs 98.88% of the time and C-13 occurs 1.108% of t
Veseljchak [2.6K]

Answer:

c) (12×0.9889) + (13×0.01108)

Explanation:

Given data:

Percentage of C-12 = 98.89%

Percentage of C-13 = 1.108%

Atomic mass = ?

Solution:

98.89/100 = 0.9889

1.108/ 100 = 0.01108

Atomic mass = (12×0.9889) + (13×0.01108)

Atomic mass = (11.8668 + 0.144034)

Atomic mass = 12.01084

6 0
3 years ago
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