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Dmitrij [34]
3 years ago
15

Convert 7120 mL to the unit dm3.

Chemistry
1 answer:
uysha [10]3 years ago
4 0
1dm^3 = 1000ml
Thus, 7120ml = 7120/1000 = 7.120dm^3
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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:

<u>V₂ = 45.0 L</u>

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
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