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3 years ago

What is the concentration of each ion in 15.00 ml of a 7.85 X 10^-6 M solution of Tc3 (PO4)7?

Chemistry

1 answer:

Daniel [21]3 years ago

5 0

Answer:

Tc^(+7) - 2.35*10^-5M

PO4(^-3) - 5.50*10^-5M

Explanation:

Tc3(PO4)7 <---> $3Tc^{7+} + 7PO_{4}^{3-}$

1 mol 3 mol 7 mol

7.85 X 10^-6 M 3*7.85*10^-6M 7*7.85*10^-6M

2.35*10^-5M 5.50*10^-5M

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Answer:

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3 years ago

some components of ink are minimally attracted to the stationary phase and very soluble in the solvent. where are these componen

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2 years ago

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the volume of liquid can be measured with? A. graduated cylinder B. pan balance C. scale D. metric ruler

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3 years ago

Calculate the pH of a 0.10 M HCN solution that is 0.0070% ionized.

Anastaziya [24]

Answer:

D) 5.15

Explanation:

Step 1: Write the equation for the dissociation of HCN

HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)

Step 2: Calculate [H⁺] at equilibrium

The percent of ionization (α%) is equal to the concentration of one ion at the equilibrium divided by the initial concentration of the acid times 100%.

α% = [H⁺]eq / [HCN]₀ × 100%

[H⁺]eq = α%/100% × [HCN]₀

[H⁺]eq = 0.0070%/100% × 0.10 M

[H⁺]eq = 7.0 × 10⁻⁶ M

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3 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

7 0

2 years ago