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SCORPION-xisa [38]

1 year ago

10

What type of chemical reaction will Ca LiOH → make

1 answer:

cricket20 [7]1 year ago

3 0

Answer:

It is not possible to determine the type of chemical reaction that will occur when Ca LiOH → without additional information. The chemical formula Ca LiOH could represent a compound, but without knowing what reactants are present and what products are being formed, it is not possible to classify the reaction. Some possible reactions that could occur involving Ca LiOH include a synthesis reaction, where Ca LiOH is formed from its constituent elements, a decomposition reaction, where Ca LiOH breaks down into its constituent elements, or a substitution reaction, where one or more atoms in Ca LiOH are replaced by other atoms.

Explanation:

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Is this equation balanced or unbalanced?<br> C+02=CO2

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This equation is balanced

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3 years ago

What is the boiling point of a solution containing 203 g of ethylene glycol (C2H6O2) and 1035 g of water? (Kb for water is 0.52

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Answer:

101,37°C

Explanation:

Boiling point elevation is one of the colligative properties of matter. The formula is:

ΔT = kb×m <em>(1)</em>

Where:

ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-

kb is ebulloscopic constant (0,52°C/m)

And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):

203g × (1mol /62,07g) = <em>3,27moles of ethlyene glycol</em>

<em />

Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m

Replacing these values in (1):

X - 100°C = 0,52°C/m×2,64m

X - 100°C = 1,37°C

<em>X = 101,37°C</em>

<em></em>

I hope it helps!

7 0

3 years ago

The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?

Agata [3.3K]

<u>Answer:</u> The molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

$PbI_2\rightleftharpoons Pb^{2+}+2I^-$

s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Pb^{2+}][I^-]^2$

We are given:

$K_{sp}=7.9\times 10^{-9}$

Putting values in above equation, we get:

$7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L$

Hence, the molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

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