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3 years ago

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If cacl2 is dissolved in water, what can be said about the concentration of the ca2+ ion? view available hint(s) if is dissolved

in water, what can be said about the concentration of the ion? it has the same concentration as the cl− ion. its concentration is half that of the cl− ion. its concentration is twice that of the cl− ion. its concentration is one-third that of the cl− ion.

2 answers:

Simora [160]3 years ago

8 0

The concentration of Ca2+ ions is half that of the Cl- ions.

Furkat [3]3 years ago

6 0

<u>Answer:</u> The correct answer is its concentration is half that of the $Cl^-$ ion.

<u>Explanation:</u>

We are given a chemical compound having formula $CaCl_2$

It is an ionic compound because it is formed by the combination of $Ca^{2+}\text{ and }Cl^-$ ion

When an ionic compound is dissolved in water, it breaks into its ions. The chemical equation for the ionization of calcium chloride follows:

$CaCl_2\rightarrow Ca^{2+}+2Cl^-$

By stoichiometry of the reaction:

1 mole of calcium chloride produces 1 mole of calcium ions and 2 moles of chloride ions

From above, it is visible that number of chloride ions is twice that of calcium ion and so will be its concentration.

Thus, the concentration of calcium ions will be half that of chloride ions.

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What is highly reactive to water and forms an ionic bond with chlorine? It’s very shiny.

34kurt

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3 years ago

Pentane is a small liquid hydrocarbon, between propane and gasoline in size at C5H12. The density of pentane is 0.626 g/mL and i

Goshia [24]

Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and $3.073\times 10^4kJ/L$ respectively.

Explanation :

Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).

As we are given that:

$\Delta H^o_{comb}=-3535kJ/mol$

Fuel value = $\frac{\Delta H^o_{comb}}{\text{Molar mass of pentane}}$

Molar mass of pentane = 72 g/mol

Fuel value = $\frac{3535kJ/mol}{72g/mol}$

Fuel value = 49.09 kJ/g

Now we have to calculate the fuel density of pentane.

Fuel density = Fuel value × Density

Fuel density = (49.09 kJ/g) × (0.626g/mL)

Fuel density = 30.73 kJ/mL = $3.073\times 10^4kJ/L$

Thus, the fuel density of pentane is $3.073\times 10^4kJ/L$

4 0

3 years ago

What is the percent concentration of a 250 ml solution with 0.120 g nacl?

Effectus [21]

0.048 percent is the answer

3 0

3 years ago

How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield

ahrayia [7]

<span>Answer: 100 ml
</span>

<span>Explanation:

1) Convert 1.38 g of Fe₂S₃ into number of moles, n

</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>

iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
</span>

<span>2) Use the percent yield to calculate the theoretical amount:
</span>

<span>65% = 0.65 = actual yield/ theoretical yield =>

</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
</span>

<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)

4) Stoichiometrical mole ratios:
</span>

<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl

5) Proportionality:

</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃

6) convert 0.020 mol to volume
</span>

<span>i) Molarity formula: M = n / V
</span>

<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>

3 0

3 years ago

Read 2 more answers

A urine sample has a mass of 147 g and a density of 1.020 g/mL. What is the volume of this sample?

son4ous [18]

So we know that the equation for density is:

$D=\frac{m}{V}$

where D is the density, m is the mass in grams, and V is the volume in mL.

So since we know two of the variables, mass and density, we can solve for the volume:

$\frac{1.020g}{mL}=\frac{147g}{V}$

$V=\frac{147g}{\frac{1.020g}{mL}}$

$V=144.12mL$

Therefore, the volume of this urine sample is 144.12mL.

5 0

3 years ago