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3 years ago

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Cho recorded the weather in her town for a week. She wrote down her observations in the table below. Mon Tue Wed Thu Fri Sat Sun

High Temperature (degrees C) 23 22 22 21 17 14 14 Amount of Precipitation (cm) 0 0 0 0 1 3 5 If the trend in the weather pattern continues over the next week, what kind of weather do you predict Cho's town will have next week?

1 answer:

SIZIF [17.4K]3 years ago

6 0

Answer:

If this trend continues, the following week will be cooler, and a large amount of rain will fall.

Explanation:

Patterns and trends can often be found in data sets. During the week that Cho recorded the weather, the temperatures consistently dropped by one to four degrees each day. At the end of the week, the amount of precipitation increased daily.

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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

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

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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,

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

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

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

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

Initial: 0.0003

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

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

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

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

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

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Answer:

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Explanation:

Given parameters:

Number of molecules of water = 1.2 x 10²¹ molecules

Unknown:

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Ensure that the equation given is balanced;

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