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2 years ago

11

Which of the following will ammonium ion form an insoluble salt? 4/16 a. chloride b. sulfate c. sulfate and carbonate d. carbona

te e. none of the above

1 answer:

Darina [25.2K]2 years ago

6 0

Answer:

e

Explanation:

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Write the balanced Ka and Kb reactions for HSO3- in water. Be sure to include the physical states of each species involved in th

Hunter-Best [27]

Answer:

Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]

Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]

Explanation:

An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:

HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)

And Ka, the acid dissociation constant is:

<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />

When base:

HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)

And kb, base dissociation constant is:

<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>

6 0

1 year ago

6.0 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 13

vodomira [7]

<u>Answer:</u>

<em>The molecular formula of X is given as $C_7 H_6 O_3$ </em>

<em></em>

<u>Explanation:</u>

$Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$$

$mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$$

mass O = Total mass of the compound - (mass of C + mass of H)

=6.0 g - ( 3.65 + 0.262 ) g

=2.09 g

$moles $O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$$

Least moles is for O that is 0.131mol and dividing all by the least we get

$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$

Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3

$\\$C 2.3 \times 3=7$\\\\$H 2 \times 3=6$\\\\$O 1 \times 3=3$$

So the empirical formula is $C_7 H_6 O_3$

Empirical formula mass

$=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g$

$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$

Molecular formula =n × empirical formula

$=1 \times C_7 H_6 O_3$

Compound X = $C_7 H_6 O_3$ is the Answer

8 0

2 years ago

Which of these facts best illustrates why regulation of alcohol consumption is necessary

exis [7]

So that people don’t break laws or drive under the influence.

4 0

2 years ago

Which of the following concerning electrolytes and the solubility rules is/are true? 1. The solubility rules apply only to ionic

Mkey [24]

Answer:

1. is true

Explanation:

The solubility rules apply only to salts, which are ionic compounds.

2. is false. A strong electrolyte is a salt that dissociates completely in solution. Not all salts dissociate completely. For example, a 0.36 mol·L⁻¹ solution dissociates as:

K₂SO₄ ⟶ K⁺ + KSO₄⁻ (30 %) + SO₄²⁻

Thus, K₂SO₄ does not dissociate completely into K⁺ and SO₄²⁻ ions.

3. is false. The solubility rules apply only to aqueous solutions.

7 0

2 years ago

How much energy is required to melt a 20.0lb bag of ice a 0°c? A pound (lb.) is equivalent to 0.4536. The Hfusion of ice is +6.0

erma4kov [3.2K]

To compute the energy, Q, needed to melt a certain amount of ice, n, in moles, we have

$Q = n \Delta H_{f}$

where the latent heat of fusion for ice is equal to 6.009 kJ/mol.

Now, since we have a 20.0 lb ice, we must first convert its mass to grams. Thus mass = (20.0)(1000)(0.4536) = 9072 g.

To find the number of moles present, we must recall that the molar mass of water (ice) is equal to <span>1.00794(2) + </span><span>15.9994 </span>≈ 18.01 g/mol. Hence, we have

$ice_{moles} = (\frac{9072 g}{1})(\frac{1 mol}{18.01 g}) =503.72 moles$

Now, to compute for the molar heat of fusion, Q,

$Q = (503.72)(6.009) = 3026.9 kJ$

Therefore, the amount of heat needed to melt the 20-lb bag of ice is equal to 3026.9 kJ.

Answer: 3026.9 kJ

6 0

2 years ago