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Nesterboy [21]
3 years ago
14

Calculate the hydronium ion concentration in an aqueous solution with a pH of 11.7 at 25°C.

Chemistry
1 answer:
DENIUS [597]3 years ago
6 0

Answer:

Approximately 2\times 10^{-12}\;\rm mol \cdot L^{-1}.

Explanation:

The hydronium ion concentration \left[\mathrm{H_3O^{+}}\right] of an aqueous solution can be found from its \rm pH with the equation:

\displaystyle \left[\mathrm{H_3O^{+}}\right] = 10^{-\mathrm{pH}}.

For this solution, \rm pH= 11.7. Hence,

\begin{aligned}& \left[\mathrm{H_3O^{+}}\right] \\ &= 10^{-\mathrm{pH}} \\ &= 10^{-11.7} \approx 2 \times 10^{-12}\end{aligned}.

Note that for this equation, the number of significant figures in \left[\mathrm{H_3O^{+}}\right] should be the same as the number of decimal places in \rm pH. For example, the \rm pH of this question comes with only one decimal place. As a result, there would be only one significant figure in the \left[\mathrm{H_3O^{+}}\right] obtained from the equation.

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if a lead can be extrated with 92.5% efficiency,what is the mass of ore required to make a lead sphere with 750%cm radius
Lady_Fox [76]
The mass of ore required is 21 700 t.

r = 750 cm

V = \frac{4}{3}  \pi  r^{3} = \frac{4}{3}  \pi  (750 cm)^{3} = 1.767 × 10⁹ cm³

The density of lead is 11.34 g/cm³.

So mass of lead sphere = 1.767 × 10⁹ cm³ × \frac{11.34 g}{1 cm^{3} } = 2.004 ×10¹⁰ g

2.004 ×10¹⁰ g × \frac{1 kg}{1000 g} = 2.004 × 10⁷ kg

2.004 × 10⁷ kg × \frac{1 t}{1000 kg} = 2.004 × 10⁴ t

92.5% efficiency means 92.5 t Pb per 100 t of ore.

Mass of ore = 2.004 × 10⁴ t Pb ×\frac{100 tore}{92.5 t Pb} = 2.17 × 10⁴ t ore = 21 700 t ore

6 0
3 years ago
A student used 10 mL water instead of 30mL for the extraction of the NaCl from the mixture. How would this affect the calculated
Usimov [2.4K]
The amount of Nacl would be 3 times more
7 0
2 years ago
Consider the following reaction where K. = 154 at 298 K: 2NO(g) + Brz(9) 2NOBr(g) A reaction mixture was found to contain 2.69x1
bekas [8.4K]

Explanation:

2NO(g) + Br_2(g)\rightleftharpoons 2NOBr(g)

Equilibrium constant of reaction = K=154

Concentration of NO = [NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M

Concentration of bromine gas = [Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M

Concentration of NOBr gas = [Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M

The reaction quotient is given as:

Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}

Q=328.06

Q>K

The reaction will go in backward direction in order to achieve an equilibrium state.

1. In order to reach equilibrium NOBr (g) must be produced.  False

2. In order to reach equilibrium K must decrease. False

3. In order to reach equilibrium NO must be produced. True

4. Q. is less than K . False

5. The reaction is at equilibrium. No further reaction will occur. False

8 0
2 years ago
Who knows how to do this please help
Fofino [41]

Answer: You forgot to zero the balance

Explanation:

5 0
2 years ago
The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O represents an acid-base tritration. The equivalence point occurs when 24.75 mL o
artcher [175]
            moles NaOH = c · V = 0.2432 mmol/mL · 24.75 mL = 6.0192 mmol
            moles H2SO4 = 6.0192 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.0096 mmol
Hence
            [H2SO4]= n/V = 3.0096 mmol / 38.94 mL = 0.07729 M
The answer to this question is  [H2SO4] = 0.07729 M

6 0
2 years ago
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