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Svet_ta [14]
3 years ago
8

.formular for Charles' law​

Chemistry
1 answer:
irga5000 [103]3 years ago
3 0

Explanation:

Hey there!

Charles law states that the volume of gas is directly proportional to the absolute temperature at constant pressure.

As the fact is, at a constant pressure the volume of fix amount of dry gas is directly proportional to absolute temperature from Charles law.

Since V and T are directly varying directly, we can write the equation using constant "k".

V/T = k

In this case the value of k depends on the pressure of gas, the amount of gas and also unit volume.

V/T = k .........(i)

Let us consider V1 and T1 the Volume and temperature of the ideal gases.

Then the equation is;

V1/T1 =k........(ii)

After this let change the temperature and volume be T2 and V2, respectively.

Then the equation is:

V2/T2 = k.......(iii)

Now; Equating equation (ii) and (iii)…

V1/T1 = V2/T2

So, this the formula. (i.e V1/T1 = V2/T2).

<u>Hope</u><u> it</u><u> helps</u><u>!</u>

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Answer:

[H+] = 1.66 x 10^{-4}

Explanation:

To find the [H+] concentration of a solution, we can use the formula:

[H+] = 10^{-pH}

Let's plug in the pH.

[H+] = 10^{-3.78}

Evaluate the exponent.

[H+] = 1.66 x 10^{-4}

Hope this helps!

6 0
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In order for them to work, what must the control rods in a nuclear reactor contain? A. neutron-emitting material
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<h2>Answer:</h2>

The correct answer is option B which is neutron absorbing material.

<h3>Explanation:</h3>
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  • Control rods are used to control the fission reaction in nuclear reactors by absorbing neutrons.
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What does D represent?
Vitek1552 [10]

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5 0
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What type of bond will two carbon atoms form?
Cerrena [4.2K]

Answer:

Nonpolar covalent  bond

Explanation:

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6 0
4 years ago
Calculate the pH for each case in the titration of 50.0 mL of 0.230 M HClO ( aq ) 0.230 M HClO(aq) with 0.230 M KOH ( aq ) . 0.2
Vikentia [17]

Answer:

pH before addition of KOH = 4.03

pH after addition of 25 ml KOH = 7.40

pH after addition of 30 ml KOH = 7.57

pH after addition of 40 ml KOH = 8.00

pH after addition of 50 ml KOH = 10.22

pH after addition 0f 60 ml KOH = 12.3

Explanation:

pH of each case in the titration given below

(6) After addition of 60 ml KOH

Since addition of 10 ml extra KOH is added after netralisation point.

Concentration of solution after addition 60 ml KOH is calculated by

M₁V₁ = M₂V₂

or, 0.23 x 10 = (50 + 60)ml x M₂

or M₂ = 0.03 Molar

so, concentration of KOH = 0.03 molar

                               [OH⁻] = 0.03 molar

                                 pOH = 0.657

                                  pH = 14 - 0.657 = 13.34

5 0
3 years ago
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