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kupik [55]
2 years ago
11

Which statement correctly pairs the climate factor with its effect on temperature? *

Chemistry
1 answer:
allochka39001 [22]2 years ago
5 0

Answer:

A region on top of a mountain is cooler than at the base.

Explanation:

Pressure and temperature have direct relationship with each other. With the decrease in pressure, the temperature decreases and vice versa. When the air rises in the atmosphere, the pressure starts to fall. The low pressure at the peak of the mountains tends to cause the fall in temperature. It is because of this reason that it is cooler at the top of the mountain while the temperature is less cool in the foothills.

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Breaker A

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Because the temperature is cooler in A then the rest

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A substance freezes at -58 celcius. Therefore a substance melts at: ________________
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-58 °C
The melting point is the same as the freezing point.
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Z forms chloride compounds with the formulae ZCl2 and ZC13.
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(B) II, IV.

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You are making homemade ice cream and can use either 200 g Ice Melt (CaCl2) or 200 g rock salt (NaCl) to lower the freezing poin
mash [69]

Answer:

Calcium chloride (CaCl2) would help me eat my ice cream faster

Explanation:

Addition of salt to ice melts the lowers the freezing temperature of the ice thus melting the ice easily

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5 0
3 years ago
Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE
nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let "\alpha is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

                  HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}

Initial                  C                         0            0

Equilibrium        c(1- \alpha)              c\alpha c\alpha

Dissociation constant = Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}

=\frac{c\alpha^{2}}{(1-\alpha)}

In this case weak acids \alpha is very small so, (1-\alpha ) is taken as 1.

Ka=C\alpha^{2}

\alpha=\sqrt\frac{ka}{c}

From the given the concentration = 0.050 M

Substitute the given value.

\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}

[H_{3}O^{+}]=c\alpha

[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}

pH= -log[H_{3}O^{+}]

=-log[4.9\times10^{-6}]

=6-log 4.9= 5.31

Therefore, The pH of the solution is 5.31.

7 0
3 years ago
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