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3 years ago

If a gas has a volume of 15l and a temperature of 125K, and the temperature is increased to 225K, what is the new volume

Chemistry

1 answer:

aalyn [17]3 years ago

6 0

Answer:

V₂ = 27 L

Explanation:

Given data:

Initial volume = 15 L

Initial temperature = 125 K

Final temperature = 225 K

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 15 L × 225K / 125 k

V₂ = 3375 L.K / 125 K

V₂ = 27 L

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How many moles of al does 13.49 g of al represent?

Harlamova29_29 [7]

Use equation
number of moles= sample mass/molar mass

7 0

3 years ago

8. What is the volume of 5.6 moles of carbon dioxide gas at STP?

pshichka [43]

Answer:

125.5 ×10^-3 m^3= 0.1255 m^3

Explanation:

Volume=5.6mol×22.414dm^3

=125.5dm^3

7 0

3 years ago

A chemistry student weighs out of sulfurous acid , a diprotic acid, into a volumetric flask and dilutes to the mark with distill

nadezda [96]

The question is incomplete, here is the complete question:

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of NaOH needed is 36.2 mL

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

$\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M$

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL$

Hence, the volume of NaOH needed is 36.2 mL

5 0

3 years ago

Which term names elements that are usually shiny and good conductors of heat and electricity?

Sonbull [250]

Elements that are usually shiny and good conductors of heat and electricity are metals.

<h3>What are conductors?</h3>

The conductors are the metal or non metal elements which allow heat and electricity to easily pass through them.

Metals are elements with good conductivity of electric current and heat. Metals have lustre characteristics. They are shiny and bendable just like copper wire, which is a metal.

Hence, elements that are usually shiny and good conductors of heat and electricity are metals.

Learn more about conductors.

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4 0

2 years ago

The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half life of 68 minutes. Suppose

viva [34]

Answer:

The concentration would be; 0.0038 μgmL

Explanation:

Half life, t1/2 = 68 minutes

Initial Conc. [A]o = 0.12/μgmL

Final Conc [A] = ?

Time. k = 340 minutes

ln[A] = ln[A]o - kt

t1/2 = ln2 / k

k = 0.693 / t1/2 = 0.693 / 68 = 0.01019

ln[A] = ln (0.12) - 0.01019 (340)

ln[A] = -2.1203 -3.4646

ln[A] = -5.5849

[A] = 0.00375 ≈ 0.0038 μgmL

3 0

3 years ago