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3 years ago

Calculate the hydroxide ion consentration of a solution with pH = 3.25. show all calculations leading to your answer.

1 answer:

6 0

First you calculate the pOH of the solution:

pH+ pOH = 14

3.25 + pOH = 14

pOH = 14 - 3.25

pOH = 10.75

Concentration of [OH]⁻ in solution:

[ OH⁻ ] = $10 ^{-pOH}$

[ OH⁻ ] = 10^ - 10.75

[OH⁻] = 1.778 x 10⁻¹¹ M

hope this helps !

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Magnesium burns in air with a dazzling brilliance to produce magnesium oxide: 2Mg (s) + O2 (g) →→ 2MgO (s) When 4.50 g of magnes

Klio2033 [76]

Answer:

7.46 g

Explanation:

From the balanced equation, 2 moles of Mg is required for 2 moles of MgO.

The mole ratio is 1:1

mole = mass/molar mass

mole of 4.50 g Mg = 4.50/24.3 = 0.185 mole

0.185 mole Mg will tiled 0.185 MgO

Hence, theoretical yield of MgO in g

mass = mole x molar mass

0.185 x 40.3 = 7.46 g

6 0

4 years ago

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How many joules are absorbed when 64.3 grams of water is heated from 40.0 °C to 74.0 °C?

Alinara [238K]

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3 0

3 years ago

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The graph below shows how solubility changes with temperature.

Otrada [13]

The two solutions that have similar solubilities at 40°C will be Na₂HAsO₄ and Na₂SO₄.

<h3>What is solubity ?</h3>

It is the ability of a solute to be dissolved, especially in water.

As the temperature increases the solubility of the gas generally decreases"

According to given information and graph attached as reference ;

Solubility of NaCl (sodium chloride) at 40°C is 36.3 grams.
Solubility of Na₂SO₄ (sodium sulfate) at 40°C is 48.8 grams.
Solubility of Na₂HAsO₄ (sodium arsenate dibasic) at 40°C is 48.9
grams.
Solubility of Ba(NO₃)₂ (barium nitrate) at 40°C is 14.1 grams.
Solubility of Ce₂(SO₄)₃·9H₂O (ceasium sulfate nonahydrate) at 40°C is 6.2 grams.

Lear more about solubility here ;

brainly.com/question/19221092

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4 0

2 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant at for the following reaction. N2(g)H

soldier1979 [14.2K]

Answer: The equilibrium constant for this reaction is $5.85\times 10^{5}$

Explanation:

The equation used to calculate standard Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]$

For the given chemical reaction:

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

The equation for the standard Gibbs free change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_{(NH_3(g))})]-[(1\times \Delta G^o_{(N_2)})+(3\times \Delta G^o_{(H_2)})]$

We are given:

$\Delta G^o_{(NH_3(g))}=-16.45kJ/mol\\\Delta G^o_{(N_2)}=0kJ/mol\\\Delta G^o_{(H_2)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol$

To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -32.9 kJ/mol = -35900 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = $25^oC=[273+25]K=298K$

$K_{eq}$ = equilibrium constant at 25°C = ?

Putting values in above equation, we get:

$-32900J/mol=-(8.314J/Kmol)\times 298K\times \ln K_{eq}\\\\K_{eq}=e^{13.279}=5.85\times 10^{5}$

Hence, the equilibrium constant for this reaction is $5.85\times 10^{5}$

5 0

3 years ago

If 7.400 g of C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final temp

Andru [333]

In your problems the possible solution is this one:

Ffr=μkFN=μkmg0.5(50000kg)(−9.8ms2)=−245000N a=F/m−245000N/50000kg=−4.9ms2t=v−v0a0−10ms−4.9ms2=2s

I hope you understand and satisfied with my answer

8 0

3 years ago