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3 years ago

V1T2 = V2T1 is an expression of who’s law

Chemistry

2 answers:

mihalych1998 [28]3 years ago

8 0

Answer:

It is an expression of Charle's law.

Explanation:

Charles' law:

It is one of the gas law.

It states that at constant pressure and for a given moles of a gas, the volume of the gas is directly proportional to the temperature (K).

If we increase the temperature of an ideal gas then at constant pressure and mole, the volume will increase linearly.

V α T

for two volumes at two different temperature

V₁ α T₁

And

V₂ α T₂

diving the two

V₁/V₂ = T₁/T₂

V₁T₂=V₂T1

Andru [333]3 years ago

6 0

Answer:

It is the expression of Charles' Law.

Explanation:

The given expression V1T2 = V2T1 is the formula for the Charles' Law.

A special case of an ideal gas is named as the Charles' Law. This law applies to ideal gases only which are at constant pressure.

According to this law, the volume of a fixed mass of a gas is directly proportional to its temperature and is given by:

V1T2 = V2T1

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Read 2 more answers

Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration

gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, $K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}$

$K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}$

Let us assume that the solubility be "s". And, the reaction equation is as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$9.84 \times 10^{-21} = s \times s$

s = $9.92 \times 10^{-11}$

Also, $Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}$

$2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}$

s = $7.14 \times 10^{-7}$

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the $K_{sp}$ expression as follows.

$K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}$

$[PO^{3-}_{4}]^{2}$ = $4.88 \times 10^{-24}$

= $2.21 \times 10^{-12}$ M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$K_{sp} = [Al^{3+}][PO^{3-}_{4}]$

$9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}$

$[Al^{3+}] = 4.45 \times 10^{-9}$ M

Thus, we can conclude that concentration of aluminium will be $4.45 \times 10^{-9}$ M when calcium begins to precipitate.

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3 years ago