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AVprozaik [17]
3 years ago
12

What is the name of the average weather in an area over a long period of time?

Chemistry
1 answer:
Nezavi [6.7K]3 years ago
3 0

Climate is the weather that occur over a long period in a particular place.

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Which one of the following substances would be the most soluble in CCl4?
maria [59]

Answer: Option (d) is the correct answer.

Explanation:

Carbon tetrachloride (CCl_{4}) is a non-polar solvent. Whereas out of the given options, Na_{2}SO_{4}, H_{2}O, CH_{3}CH_{2}CH_{2}CH_{2}OH, and HI are all polar molecules.

On the other hand, only C_{4}H_{10} is non-polar molecule.

Also it is known that like dissolves like.

So, being non-polar CCl_{4} will dissolve the give alkane, C_{4}H_{10}.

5 0
3 years ago
A balloon has a volume of 125mL with 10 pumps of gas. The balloon is reduced in volume to 88mL, how much gas is in the balloon?
blagie [28]

Answer:

7.04

Explanation:

i think

6 0
3 years ago
How many grams of CaCl 2 are in .48 moles ?
Llana [10]

Answer:

We assume you are converting between moles CaCl2 and gram. You can view more details on each measurement unit: molecular weight of CaCl2 or grams This compound is also known as Calcium Chloride. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles CaCl2, or 110.984 grams.

7 0
2 years ago
Read 2 more answers
The 10 short columns in the middle of the periodic table make up the
nekit [7.7K]

Answer:

The transition elements or transition metals occupy the short columns in the center of the periodic table, between Group 2A and Group 3A.Explanation:

6 0
2 years ago
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What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
3 years ago
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