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Tema [17]
3 years ago
14

Liquid nitrogen has a density of 0.807 g/ml at –195.8 °c. if 1.00 l of n2(l) is allowed to warm to 25°c at a pressure of 1.00 at

m, what volume will the gas occupy? (r = 0.08206 l×atm/k×mol)
Chemistry
1 answer:
Flauer [41]3 years ago
8 0
Step 1: Change density from g/mL to g/L;

                                  0.807 g/mL  =  807 g/L

Step 2: Find Moles of N₂;
As,
             Density  =  Mass / Volume
Or,
             Mass  =  Density × Volume

Putting Values,

            Mass  =  807 g/L × 1 L

            Mass  =  807 g
Also,
            Moles  =  Mass / M.mass

Putting values,

            Moles  =  807 g / 28 g.mol⁻¹

            Moles  =  28.82 moles

Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,

As,
                        P V  =  n R T

                            V  =  n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)

                 V  =  (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm

                 V  =  704.76 L
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2 years ago
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How much water must be added to 6.0 M silver nitrate in order to make 500 mL of 1.2 M solution?
algol [13]

The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL

<u>Given data :</u>

Concentration of siilver nitrate ( M₁ ) = 6.0 M

volume of solution ( V₁ ) = 500 mL

Conc of solution ( M₂ )= 1.2 M

<h3>Determine the amount of water that must be added</h3>

we will apply the equation below

M₁V₁ = M₂V₂ ---- ( 1 )

where : V₂ = V₁ + water added  ---- ( 2 )

V₂ ( Final volume ) = ( M₁V₁ ) / M₂

                              = ( 6 * 500 ) / 1.2

                              = 2500 mL

Back to eqaution ( 2 )

2500 mL = 500 mL + added water

therefore ; added water = 2500 - 500

                                        = 2000 mL

Hence we can conclude that The amount of water that must be added to 6.0 M silver nitrate to make 500mL of 1.2 M solution is : 2000 mL.

Learn more about Volume : brainly.com/question/12410983

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7 0
2 years ago
a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re
Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n}{V_s}

where,

n = moles of solute

V_s = volume of solution in L

Given : 59.4 g of H_2SO_4 in 100 g of solution  

moles of H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61

Volume of solution =\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml

Now put all the given values in the formula of molality, we get

Molality=\frac{0.61\times 1000}{54.6ml}=11.2M

To calculate the volume of acid, we use the equation given by neutralisation reaction:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock acid which is H_2SO_4

M_2\text{ and }V_2 are the molarity and volume of dilute acid which is H_2SO_4

We are given:

M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL

Putting values in above equation, we get:

11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0
3 years ago
We saw in the calorimeter experiment that the combustion of a fuel is exothermic. what can you then say about the formation of a
ankoles [38]

Answer:

This question appear incomplete

Explanation:

This question appear incomplete. However, fuel is formed through a natural phenomenon involving the conversion of large amount dead and decayed organisms (usually algae and zooplanktons) to combustible fuel through exposure to relatively high temperature and pressure (over millions of years) in the earth's crust. Thus, since this involves a sort of absorption of heat energy (from the earth's crust), it can be referred to be an endothermic reaction.

8 0
2 years ago
What does the symbol Cp
xenn [34]

Answer:

From what i've learned so far, the correct answer is "Heat at a constant Pressure" or "Specific Heat"

Explanation:

Hope this helps!

8 0
2 years ago
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