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3 years ago

What’s the answer? I need an answer as soon as possible please!

Chemistry

2 answers:

inysia [295]3 years ago

5 0

Answer:

1, 1, 2

Explanation:

1, 1, 2

balandron [24]3 years ago

4 0

N2O5 + H2O → 2HNO3

The coefficients for the N2O5 and H2O are technically 1; if you must provide a value for those reactants, then that would be it.

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The density of ammonia gas under certain conditions is 0.625 g/l. calculate its density in g/cm3.

MatroZZZ [7]

From the conversion of units:
1 cm^3 is equivalent to 1 mL
1 L is equivalent to 1000 mL
therefore,
to convert from liter to cm^3, we simply multiply by 1000.
Note that the multiplication will be done in the denominator.

Based on this:
density = (0.625 g/l) x (1g/1000 cm^3) = <span> 0.000625 g/cm^3
= 6.25 x 10^-4 g/cm^3</span>

6 0

3 years ago

A gas of 3.4 moles occupies a volume of 0.046 L at 298 K. What is the pressure in kPa?

Phantasy [73]

Answer : The correct option is, $2.1\times 10^5kPa$

Explanation :

To calculate the pressure of gas we are using ideal gas equation as:

$PV=nRT$

where,

P = pressure of gas = ?

V = volume of gas = 0.046 L

n = number of moles of gas = 3.4

R = gas constant = 8.314 L.kPa/mol.K

T = temperature of gas = 298 K

Now put all the given values in the above formula, we get:

$P\times (0.046L)=(3.4mol)\times (8.314L.kPa/mol.K)\times (298K)$

$P=1.83\times 10^5kPa\approx 2.1\times 10^5kPa$

Therefore, the pressure of gas is, $2.1\times 10^5kPa$

3 0

3 years ago

PLEASE HELP!

Musya8 [376]

1) Decomposition

2) CaCO3 → CaO + CO2

equation is already balanced

6 0

3 years ago

What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 3.0 ATM and the temperatu

Arisa [49]

Answer:

$\boxed {\boxed {\sf P_2= 9 \ atm}}$

Explanation:

Since the temperature is held constant, we only need to focus on the volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula is:

$P_1V_1=P_2V_2$

Originally, the gas had a volume of 150 milliliters and a pressure of 3.0 atmospheres. We can substitute these values into the left side of the equation.

$3.0 \ atm * 150 \ mL = P_2V_2$

The original gas was compressed to a volume of 50 milliliters, but we don't know the volume.

$3.0 \ atm *150 \ mL= P_2 * 50 \ mL$

Now, we need to solve for the new pressure (P₂). Multiply on the left side first.

$450 \ atm*mL= P_2 * 50 \ mL$

Since we are solving for the pressure, we need to isolate the variable. It is being multiplied by 50 mL. The inverse of multiplication is division. Divide both sides by 50 mL.

$\frac{450 \ atm*mL}{50 \ mL}= \frac{P_2 * 50 \ mL}{50 \ mL}$

$\frac{450 \ atm*mL}{50 \ mL}= P_2$

The units of milliliters will cancel.

$\frac{450 \ atm}{50 }= P_2$

$9 \ atm =P_2$

The new pressure is <u>9 atmospheres.</u>

8 0

3 years ago

A helium gas balloon is expanded to 78.0 L, while the pressure is held constant at 0.37 atm. If the work done on the gas mixture

Elenna [48]

Answer:

77.248 L

Explanation:

From the question,

Work done on the gas mixture is given as,

W = PΔV.................. Equation 1

Where W = work done, P = pressure of the the gas, ΔV = Change in volume of the gas.

make ΔV the subject of the equation

ΔV = W/P..................... Equation 2

Given: W = 28.2 J, P = 0.37 atm = (0.37×101325) N/m² = 37490.25 N/m²

Substitute into equation 2

ΔV = 28.2/37490.25

ΔV = 0.000752 m³

ΔV = 0.752 L

But,

ΔV = V₂-V₁................. Equation 3

Where V₂ = Final volume of the helium gas, V₁ = Initial volume of the helium gas

make V₁ the subject of the equation

V₁ = V₂-ΔV................ Equation 4

Given: V₂ = 78 L.

Substitute into equation 4

V₁ = 78-0.752

V₁ = 77.248 L

8 0

3 years ago

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