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Elis [28]
3 years ago
9

All matter can be classified by their physical properties. Which substance can be classified as being soluble in water

Chemistry
1 answer:
My name is Ann [436]3 years ago
4 0
The substance that can be classified as being soluble in water would be salt. The solution is D. Salt specifically is composed of ions and the ionic compound dissociates within water.
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A pH scale reading 13 indicates a ______
lukranit [14]

Answer:

A pH scale reading 13 indicates a strong base.

Explanation:

From my understanding:

1 -4 is a strong acid

4 - 7 is weak acid

7 - 9 is a weak base

9 - 14 is a strong base

6 0
2 years ago
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All of the following are forms of acid precipitation EXCEPT ________.
Aneli [31]
Dihydrogen oxide is the right answer. Dihydrogen oxide is just 2 hydrogen and 1 oxygen which is H2O or water.
6 0
2 years ago
Picture this in your mind as you read it: You weigh an empty, clean, dry beaker on the balance (scale). It weighs 32.15 grams. Y
AleksAgata [21]

Answer:

2.25 g

Explanation:

The mass of the solid X must be the total mass (beaker + solid X) less than the mass of the beaker. Then:

mass of the solid X = 34.40 - 32.15

mass of the solid X = 2.25 g

The difference of 0.25 g must occur for several problems: an incorrect weight in the balance, the configuration of the balance, the solid can be hydrophilic and absorbs water, and others.

6 0
3 years ago
One of the most common household toxins, benzene can be found in the following:
maksim [4K]

Answer : Option D) Paint Supplies.

Explanation : The paint is usually the major common household toxins, which contains benzene in it. Benzene being carcinogenic in nature is very harmful to humans.

7 0
3 years ago
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A sample of gas with an initial volume of 12.5 L at a pressure of 784 torr and a temperature of 295 K is compressed to a volume
Kipish [7]

Answer:

Final pressure in (atm) (P1) = 6.642 atm

Explanation:

Given:

Initial volume of gas (V) = 12.5 L

Pressure (P) = 784 torr

Temperature (T) = 295 K

Final volume (V1) = 2.04 L

Final temperature (T1) = 310 K

Find:

Final pressure in (atm) (P1) = ?

Computation:

According to combine gas law method:

\frac{PV}{T} =\frac{P1V1}{T1} \\\\\frac{(784)(12.5)}{295} =\frac{(P1)(2.04)}{310}\\\\33.22 = \frac{(P1)(2.04)}{310}\\\\P1=5,048.18877

⇒ Final pressure (P1) = 5,048.18877 torr

⇒ Final pressure in (atm) (P1) = 5,048.18877 torr / 760

⇒ Final pressure in (atm) (P1) = 6.642 atm

3 0
3 years ago
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