Convert 7120 mL to the unit dm3.

A large balloon is initially filled to a volume of 25.0 L at 353 K and a pressure of 2575 mm Hg. What volume of gas will the bal

bija089 [108]

Answer:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 25.0 L

V₂ = ?

P₁ = 2575 mm Hg

Also, P (atm) = P (mm Hg) / 760

P₁ = 2575 / 760 atm = 3.39 atm

P₂ = 1.35 atm

T₁ = 353 K

T₂ = 253 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}$

$\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}$

Solving for V₂ , we get:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

4 0

3 years ago

3. Determine the uses of the following materials:

hichkok12 [17]

Answer:

D

Explanation:

D

3 0

3 years ago

It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural gas (principall

Aleks [24]

Answer:

Hydrogen: -141 kJ/g

Methane: -55kJ/g

The energy released per gram of hydrogen in its combustion is higher than the energy released per gram of methane in its combustion.

Explanation:

According to the law of conservation of the energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.

Qc + Qb = 0

Qc = -Qb [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Q = C . ΔT

where,

C is the heat capacity

ΔT is the change in the temperature

<h3>Hydrogen</h3>

Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (14.3°C) = -162 kJ

The heat released per gram of hydrogen is:

$\frac{-162kJ}{1.15g} =-141 kJ/g$

<h3>Methane</h3>

Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (7.3°C) = -82 kJ

The heat released per gram of methane is:

$\frac{-82kJ}{1.50g} =-55kJ/g$

3 0

3 years ago

Which of these solids is insoluble in water?

eimsori [14]

The answer would be A) sand, it is not soluble in water

5 0

3 years ago

Read 2 more answers

if 0.0203 g of a gas dissolves in 1.39 l of water at 1.02 atmospheres at what pressure (in atm) would you be able to dissolve 0.

Sergeu [11.5K]

First, we need to get n1 (no.of moles of water ): when

mass of water = 0.0203 g and the volume = 1.39 L

∴ n1 = mass / molar mass of water

= 0.0203g / 18 g/mol
= 0.00113 moles

then we need to get n2 (no of moles of water) after the mass has changed:

when the mass of water = 0.146 g

n2 = mass / molar mass

= 0.146g / 18 g/ mol
= 0.008 moles

so by using the ideal gas formula and when the volume is not changed:

So, P1/n1 = P2/n2

when we have P1 = 1.02 atm

and n1= 0.00113 moles

and n2 = 0.008 moles

so we solve for P2 and get the pressure

∴P2 = P1*n2 / n1

=1.02 atm *0.008 moles / 0.00113 moles

= 7.22 atm

∴the new pressure will be 7.22 atm

5 0

3 years ago