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3 years ago

44) How many grams are in 3.3 x1023 molecules of N213

Chemistry

1 answer:

Svetlanka [38]3 years ago

5 0

Answer:

Explanation:

Because i dont no tang ina

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14. What characteristics of Earth's surface can be determined by using a topographic map?

Leno4ka [110]

The answer is letter a. Hill terrain, hill slope, and hill elevation

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3 years ago

Volume increases, pressure stays the same, temperature

Mila [183]

Temperature decreases (?)

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3 years ago

What is the density of an object that has a volume of 34.2 cm^3 and a mass of 19.6 g? (a) 0.573 g/cm^3 (b) 1.74 g/cm^3 (c) 670 g

KengaRu [80]

Answer: The density of the object will be $0.573g/cm^3$

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of object = 19.6 grams

Volume of object= $34.2cm^3$

Putting in the values we get:

$Density=\frac{19.6g}{34.2cm^3}=0.573g/cm^3$

Thus density of the object will be $0.573g/cm^3$

8 0

4 years ago

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

emmainna [20.7K]

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.

8 0

3 years ago

Hydrocarbons, compounds containing only carbon and hydrogen, are important in fuels. The heat of combustion of cyclobutane, C4H8

Gennadij [26K]

Answer:

1. C4H8 + 6O2 -----> 4CO2 + 4H20

2. 3836.77 kcal

Explanation:

1. Balanced equation for the complete combustion of cyclobutane:

C4H8 + 6O2 -----> 4CO2 + 4H20

2. Heat of combustion of cyclobutane = 650.3 kcal/mol

Molecular weight of cyclobutane, C4H8 = 56.1 g/mol

Mole of C4H8 : mass of cyclobutane/Molecular weight of cyclobutane

Mole of C4H8 = 331/56.1 = 5.9 mol

Energy released during combustion = 5.9 mol × 650.3 kcal/mol = 3836.77kcal

Therefore the energythat is released during the complete combustion of 331 grams of cyclobutane is 3836.77kcal

8 0

3 years ago