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anyanavicka [17]

3 years ago

9

PLEASE HELPP Will mark brainliest

1 answer:

zheka24 [161]3 years ago

7 0

Answer:

B

Explanation:

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13 moles of water to molecules

FromTheMoon [43]

I’m confused what are you asking exactly?

6 0

3 years ago

Which is warmer -30C or -30 F?

Burka [1]

-30°C is more warmer than -30°F...!

<h2>How??</h2>

Always Celsius is warmer than farheniet and also in coldness Celsius is colder than farheneit.

8 0

2 years ago

Read 2 more answers

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

You have 0.5 L of air at 203 k in an expandable container at constant pressure. You heat the container to 273 k. What is the vol

Jobisdone [24]

You can use P1V1/T1 = P2V2/T2 but since pressure is constant is becomes V1/T1=V2/T2

V1=0.5 L
T1=203 K
T2=273 K
V2=unknown

0.5L/203 = V2/273
V2= 0.67 L so C

Hope this helps :)

4 0

3 years ago

A sample of argon gas at 55°C is under 845 mm Hg pressure. What will the new temperature be if the pressure is raised to 1050 mm

Maurinko [17]

Answer:

The final temperature at 1050 mmHg is 134.57 $^{\circ}C$ or 407.57 Kelvin.

Explanation:

Initial temperature = T = 55 $^{\circ}C$ = 328 K

Initial pressure = P = 845 mmHg

Assuming final to be temperature to be T' Kelvin

Final Pressure = P' = 1050 mmHg

The final temperature is obtained by following relation at constant volume

$\displaystyle \frac{P}{P'}=\displaystyle \frac{T}{T'} \\ \displaystyle \frac{845 \textrm{ mmHg}}{1050 \textrm{ mmHg}} = \displaystyle \frac{328 \textrm{ K}}{T'} \\T' = 407.57 \textrm{ Kelvin}$

The final temperature is 407.57 K

7 0

2 years ago