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icang [17]
3 years ago
13

Which expression is equivalent to (q Superscript 6 Baseline) squared?

Chemistry
2 answers:
laiz [17]3 years ago
8 0

Answer:

c

Explanation:

Artyom0805 [142]3 years ago
4 0

Answer:

The answer is c

Explanation:

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Convection which allows heat to be transferred between the air molecule via convection currents .
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3 years ago
What is the freezing point of an aqueous glucose solution that has 25.0 g of glucose?
LiRa [457]

Answer:

T_f=-2.58\°C

Explanation:

Hello,

In this case, we can compute the the freezing point depression by using the following formula:

T_f-T_0=-i*m*Kf

Whereas the freezing point of pure water is 0 °C van't Hoff factor for glucose is 1, the molality is computed as shown below and the freezing point constant of water is 1.86 °C/m:

m=\frac{25.0g\ glucose*\frac{1mol\ glucose}{180g\ glucose} }{100g*\frac{1kg}{1000g} }\\ \\m=1.39m

Thus, the freezing point of the solution is:

T_f=T_0-i*m*Kf\\\\T_f=0\°C-1*1.39m*1.86\frac{\°C}{m}\\ \\T_f=-2.58\°C

Regards.

4 0
4 years ago
Balance the equations. Plzzzzz help me!!!
Leviafan [203]

Answer:

Hci

Explanation:

6 0
4 years ago
How do hydrogen fuel cells work​
Maksim231197 [3]
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4 0
4 years ago
A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t
pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = pK_a=9.31

First we have to calculate the moles of HCN and NaCN.

\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole

and,

\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole

The balanced chemical reaction is:

                          HCN+NaOH\rightarrow NaCN+H_2O

Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

                        0.0655                       0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=9.31+\log (\frac{0.1359}{0.0655})

pH=9.63

Therefore, the pH of the solution is, 9.63

4 0
4 years ago
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