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3 years ago

Calculate the amount of heat needed to evaporate 235.0 grams of water from 25.0°C to 100.0°C.

1 answer:

6 0

I hope this helps answer your question. :)

These problems are very hard to do digitally so I would recommend trying to practice these types of problems on paper :)

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A solid mixture consists of 47.6g of KNO3 (potassium nitrate) and 8.4g of K2SO4 (potassium sulfate). The mixture is added to 130

IgorLugansk [536]

Answer: No crystals of potassium sulfate will be seen at 0°C for the given amount.

Explanation:

We are given:

Mass of potassium nitrate = 47.6 g

Mass of potassium sulfate = 8.4 g

Mass of water = 130. g

Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g

This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water

Applying unitary method:

In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams

So, in 130 grams of water, the amount of potassium sulfate dissolved will be $\frac{7.4}{100}\times 130=9.62g$

As, the soluble amount is greater than the given amount of potassium sulfate

This means that, all of potassium sulfate will be dissolved.

Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.

7 0

3 years ago

Which of these molecules is nonpolar? a.CH3CI b.CO c.O2 d.PF3

valentina_108 [34]

Answer:

i think the answer is D

Explanation:

7 0

2 years ago

Read 2 more answers

What is the ph of a peach with a [ –oh] = 3.2 × 10 –11 m?

3241004551 [841]

Answer:

Amerigo Vespucci, was the first European to reach the Caribbean Islands.

Please select the best answer from the choices provided

Explanation:

7 0

2 years ago

the chemical compound C2F4 is used to make PTEE (Teflon). How manyC2F4 molecules are in 485 kilograms of this material?

monitta

Answer:

$molecules=2.92x10^{27}moleculesC_2F_4$

Explanation:

Hello,

In this case, we use the Avogadro's number to compute the molecules of C2F4 whose molar mass is 100 g/mol contained in a 485-kg sample as shown below:

$molecules=485kgC_2F_4*\frac{1000gC_2F_4}{1kgC_2F_4} *\frac{1molC_2F_4}{100gC_2F_4}*\frac{6.022x10^{23}molecules C_2F_4}{1molC_2F_4} \\\\molecules=2.92x10^{27}moleculesC_2F_4$

Best regards,

5 0

3 years ago

Because the light rays never meet, a concave lens can produce

Arturiano [62]

Answer:

Virtual images.

Hope this helps.

4 0

3 years ago