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3 years ago

What is the substance called that is being dissolved in a solution?

Chemistry

2 answers:

Katena32 [7]3 years ago

6 0

Answer:

Solvent

Explanation:

The solvent is the substance that is being dissolved the other substance

Anna [14]3 years ago

5 0

Answer:

Solute - The solute is the substance that is being dissolved by another substance. In the example above, the salt is the solute. Solvent - The solvent is the substance that dissolves the other substance. In the example above, the water is the solvent.

Explanation:

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Cindy made tea. She started with 300 grams of water at 20 degrees Celsius. She transferred 18,000 calories to the water. What wa

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Answer:

T final = 80°C

Explanation:

Q = mCpΔT

∴ Q = 18000 cal

∴ m H2O = 300 g

∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K

∴ T1 = 20°C = 293 K

∴ T2 = ?

⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)

⇒ (18000 cal)/(300 cal/K) = T2 - 293 K

⇒ T2 = 293 K + 60 K

⇒ T2 = 353 K (80°C)

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3 years ago

A reaction generates hydrogen gas (H) as a product. The re-

Amiraneli [1.4K]

Answer: The average rate of the reaction is $7.82\times 10^{-3}M/min$

Explanation:

To calculate the molarity of hydrogen gas generated, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of hydrogen gas = $3.91\times 10^{-2}mol$

Volume of solution = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$\text{Molarity of }H_2=\frac{3.91\times 10^{-2}mol}{0.250L}=0.1564M$

Average rate of the reaction is defined as the ratio of concentration of hydrogen generated to the time taken.

To calculate the average rate of the reaction, we use the equation:

$\text{Average rate of the reaction}=\frac{\text{Concentration of hydrogen generated}}{\text{Time taken}}$

We are given:

Concentration of hydrogen generated = 0.1564 M

Time taken = 20.0 minutes

Putting values in above equation, we get:

$\text{Average rate of the reaction}=\frac{0.1564M}{20.0min}\\\\\text{Average rate of the reaction}=7.82\times 10^{-3}M/min$

Hence, the average rate of the reaction is $7.82\times 10^{-3}M/min$

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2 years ago

Convert 2.68 mol H2O to molecules H2O Timed test

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Molecules h20 is .134

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2 years ago

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2 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

2 years ago