What’s the answer to this?

Li2SO4 _____ an electrolyte in solution. A. Is B. Is not

Rasek [7]

Answer:

a

Explanation:

it is an electrolyte because of its strong polar chemical bond

8 0

3 years ago

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How do you know if a equation is balanced or unbalanced

denpristay [2]

A chemical equation is only balanced once there are equal numbers of all types of element on both sides of the equation

7 0

3 years ago

In Lewis dot diagrams, valence electrons are represented by dots around the atomic symbol. How many electrons should around an a

Margaret [11]

There should be 3 valence electrons surrounding the aluminum ion

8 0

4 years ago

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A weak monoprotic acid has molar mass 180 g/mol. When 1.00 g of this acid is dissolved in enough water to obtain a 300 mL soluti

Kaylis [27]

Answer: The value of $K_a$ for the given acid is $3.58\times 10^{-4}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Initial mass of weak monoprotic acid = 1.00 g

Molar mass of weak monoprotic acid = 180 g/mol

Volume of solution = 300 mL

Putting values in above equation, we get:

$\text{Molarity of weak monoprotic acid}=\frac{1.00\times 1000}{180\times 300}\\\\\text{Molarity of weak monoprotic acid}=0.0185M$

To calculate the hydrogen ion concentration for given pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 2.62

Putting values in above equation, we get:

$2.62=-\log[H^+]$

$[H^+]=10^{-2.62}=2.40\times 10^{-3}M=0.0024M$

The chemical equation for the dissociation of weak monoprotic acid (HA) follows:

$HA\rightleftharpoons H^++A^-$

Initial: 0.0185

At eqllm: 0.0185-x x x

Evaluating the value of 'x'

$\Rightarrow x=0.0024$

So, equilibrium concentration of HA = (0.0185 - 0.0024) = 0.0161 M

Equilibrium concentration of $A^-$ = x = 0.0024 M

The expression of $K_a$ for above equation follows:

$K_a=\frac{[H^+][A^-]}{[HA]}$

Putting values in above equation, we get:

$K_a=\frac{(0.0024)\times (0.0024)}{0.0161}=3.58\times 10^{-4}$

Hence, the value of $K_a$ for the given acid is $3.58\times 10^{-4}$

5 0

4 years ago

What is the charge of the ion sulfur would become when it has a full valence shell?

MatroZZZ [7]

Your answer to that would be A. -2

4 0

3 years ago