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monitta
4 years ago
5

What is the volume of a 0.0355mol/L calcium hydroxide solution required to react completely with 50.0mL of 0.250mol/L aluminum s

ulfate solution
Chemistry
1 answer:
Salsk061 [2.6K]4 years ago
7 0

Answer:

V₁ =  0.35 L

Explanation:

Given data:

Molarity of calcium hydroxide = 0.0355 M

Volume of aluminium sulfate = 50.0 mL ( 50/1000 = 0.05 L)

Molarity of aluminium sulfate = 0.250 M

Volume of calcium hydroxide = ?

Solution:

M₁V₁ = M₂V₂

V₁ = M₂V₂ / M₁

V₁ =  0.250 M .  0.05 L / 0.0355 M

V₁ =  0.0125 M.L/ 0.0355 M

V₁ =  0.35 L

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KIM [24]

Answer: 24 moles of SnF_2 are produced.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

{\text{Moles of}  H_2}=\frac{48g}{2g/mol}=24moles

Sn+2HF\rightarrow SnF_2+H_2

According to stoichiometry :

1 mole of H_2 is accompanied with = 1 mole of SnF_2

Thus 24 moles of H_2  is accompanied with =\frac{1}{1}\times 24=24moles  of SnF_2

Thus 24 moles of SnF_2 are produced.

8 0
3 years ago
Calculate the pH of a 0.22 M ethylamine solution.
evablogger [386]

Answer:

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Explanation:

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4 0
3 years ago
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Which one of the following will change the value of an equilibrium constant?
Ad libitum [116K]

Answer:

(E) changing temperature

Explanation:

Consider the following reversible balanced reaction:

aA+bB⇋cC+dD

If we know the molar concentrations of each of the reaction species, we can find the value of Kc using the relationship:

Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)

where:

[C] and [D] are the concentrations of the products in the equilibrium; [A] and [B] reagent concentrations in equilibrium; already; b; c and d are the stoichiometric coefficients of the balanced equation. Concentrations are commonly expressed in molarity, which has units of moles / 1

There are some important things to remember when calculating Kc:

-  <em>Kc is a constant for a specific reaction at a specific temperature</em>. If you change the reaction temperature, then Kc also changes

- Pure solids and liquids, including solvents, are not considered for equilibrium expression.

- The reaction must be balanced with the written coefficients as the minimum possible integer value in order to obtain the correct value of Kc

8 0
3 years ago
What would be the major product obtained from hydroboration–oxidation of the following alkenes?
zmey [24]

Answer:

a. 3-methylbutan-2-ol

b. 2-methylcyclohexan-1-ol

Explanation:

For this reaction, we must remember that the hydroboration is an <u>"anti-Markovnikov" reaction</u>. This means that the "OH" will be added at the <em>least substituted carbon of the double bond.</em>

In the case of <u>2-methyl-2-butene</u>, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore <u>the "OH" will be added to carbon three</u> producing <u>3-methylbutan-2-ol</u>.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore<u> the "OH" will be added to carbon 2</u> producing <u>2-methylcyclohexan-1-ol</u>.

See figure 1

I hope it helps!

8 0
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zhenek [66]

1. its temperature will rise continuously until it melts

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3 years ago
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