What is true of boric acid?<br> PLEASE HELP

If the pressure on a 1.04 L sample of gas is doubled at constant temperature, please compute the new volume of gas: __

natulia [17]

Answer:

0.52 L.

Explanation:

Let P be the initial pressure.

From the question given above, the following data were obtained:

Initial pressure (P1) = P

Initial volume (V1) = 1.04 L

Final pressure (P2) = double the initial pressure = 2P

Final volume (V2) =?

The new volume (V2) of the gas can be obtained by using the the Boyle's law equation as shown below:

P1V1 = P2V2

P × 1.04 = 2P × V2

1.04P = 2P × V2

Divide both side by 2P

V2 = 1.04P /2P

V2 = 0.52 L

Thus, the new volume of the gas is 0.52 L.

6 0

3 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

Can you give a step by step direction slime experiment?

fredd [130]

Slime is made of glue, water, and borax

5 0

3 years ago

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Classify the possible combinations of signs for a reaction's ΔH and ΔS values by the resulting spontaneity

FinnZ [79.3K]

Answer :

(A) ΔH is positive and ΔS is negative → (2) Spontaneous in reverse at all temperatures

(B) ΔH is positive and ΔS is positive → (3) Spontaneous as written above a certain temperature

(C) ΔH is negative and ΔS is positive → (1) Spontaneous as written at all temperatures

(D) ΔH negative and ΔS is negative → (4) Spontaneous as written below a certain temperature

Explanation :

According to Gibb's equation:

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibbs free energy

$\Delta H$ = enthalpy change

$\Delta S$ = entropy change

T = temperature in Kelvin

As we know that:

$\Delta G$ = +ve, reaction is non spontaneous

$\Delta G$ = -ve, reaction is spontaneous

$\Delta G$ = 0, reaction is in equilibrium

(A) ΔH is positive and ΔS is negative.

$\Delta G=\Delta H-T\Delta S$

$\Delta G=(+ve)-T(-ve)$

$\Delta G=(+ve)$

The reaction is non-spontaneous at all temperatures or spontaneous in reverse at all temperatures.

(B) ΔH is positive and ΔS is positive.

$\Delta G=\Delta H-T\Delta S$

$\Delta G=(+ve)-T(+ve)$

$\Delta G=(+ve)$ (at low temperature) (non-spontaneous)

$\Delta G=(-ve)$ (at high temperature) (spontaneous)

The reaction is spontaneous as written above a certain temperature.

(C) ΔH is negative and ΔS is positive.

$\Delta G=\Delta H-T\Delta S$

$\Delta G=(-ve)-T(+ve)$

$\Delta G=(-ve)$

The reaction is spontaneous as written at all temperatures

(D) ΔH is negative and ΔS is negative.

$\Delta G=\Delta H-T\Delta S$

$\Delta G=(-ve)-T(-ve)$

$\Delta G=(+ve)$ (at high temperature) (non-spontaneous)

$\Delta G=(-ve)$ (at low temperature) (spontaneous)

The reaction is spontaneous as written below a certain temperature.

4 0

3 years ago

The blood carries nutrients to where they are needed. True False

AfilCa [17]

Answer:

True

Explanation:

8 0

3 years ago

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What is true of boric acid? PLEASE HELP