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2 years ago

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Which would be the best way to represent the concentration of a 1. 75 M K2CrO4 solution? 1. 75% K2CrO4 [K2CrO4] (K2CrO4) K2CrO4,

[M] = 1. 75.

1 answer:

tekilochka [14]2 years ago

8 0

The amount of the solute present in the given solution is called the concentration. The best way to represent the concentration of the solution is $\rm [K_{2}CrO_{4}].$

<h3>What is molar concentration?</h3>

Molar concentration is the molarity of the solution that is the measure of the concentration of the solute dissolved in the solution.

The formula for calculating molar concentration is given as,

$\rm Molarity = \dfrac{\text{moles of solute}}{\text{Solution in L}}$

The concentration of any substance is represented in the square bracket like $\rm [X]$ or $\rm [K_{2}CrO_{4}].$

Therefore, option B. $\rm [K_{2}CrO_{4}]$ is the representation of the concentration.

Learn more about the molarity here:

brainly.com/question/1532164

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We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

I'm giving 20 points! please it's for a friend not me

astra-53 [7]

Answer:

Three chemical elements: hydrogen, oxygen, and helium.

The difference between metals and metalloids is: metalloids have properties in between those of the metals and non-metals and are semiconductors.

The periodic table is organized by the elemts atomic number, it goes from the element with the lowest atomic number (which is hydrogen) to the element with the highest atomic number (oganesson)

Explanation:

Hope this helps :)

6 0

3 years ago

A hot lump of 30.5 g of iron at an initial temperature of 52.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to r

slava [35]

Answer:

26.7°C

Explanation:

Using the formula; Q = m × c × ΔT

Where; Q = amount of heat

m = mass

c = specific heat

ΔT = change in temperature

In this question involving iron placed into water, the Qwater = Qiron

For water; m= 50g, c = 4.18 J/g°C, Initial temp= 25°C, final temp=?

For iron; m = 30.5g, c = 0.449J/g°C, Initial temp= 52.7°C, final temp=?

Qwater = -(Qiron)

m × c × ΔT (water) =- {m × c × ΔT (iron)}

50 × 4.18 × (T - 25) = - {30.5 × 0.449 × (T - 52.7)}

209 (T - 25) = - {13.6945 (T - 52.7)}

209T - 5225 = -13.6945T + 721.7

209T + 13.6945T = 5225 + 721.7

222.6945T = 5946.7

T = 5946.7/222.6945

T = 26.7

Hence, the final temperature of water and iron is 26.7°C

8 0

3 years ago

Identify the arrows that represent the process of cooling.<br> gas<br> liquid

olga nikolaevna [1]

Answer:

Liquid - Gas is Evaporation & Gas - Cooling is condensation

Explanation:

3 0

2 years ago

There are two naturally occurring isotopes of indium: indium-113 and indium-115. How many neutrons are in a single atom of indiu

KonstantinChe [14]

There are 66 neutrons in a single atom of indium-115. The atomic number of indium-115 is 49, meaning there are 49 protons. Then the atomic mass is 115, so 115-49 = 66.

5 0

3 years ago