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Taya2010 [7]
3 years ago
11

Which is not a chemical substancea. solidb. plasmac. gasd. none of the above

Chemistry
1 answer:
weeeeeb [17]3 years ago
5 0
Chemical substances can exist as solids, plasma, and gases so I guess it is none of the above.

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Which word should replace the question mark in the diagram?
shutvik [7]

Answer:

Could you provide a diagram?

Explanation:

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Identify all the examples that support the Law of Conservation of Mass through chemical changes.
wariber [46]

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B

C

Explanation:

7 0
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What is one property for each of the chemical substances listed above?
Vilka [71]
1. Acetic acid (Ethanoic acid) is a colorless liquid.

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8 0
3 years ago
How long will a current of 0.995 A need to be passed through water (containing H2SO4) for 5.00 L of O2 to be produced at STP
DIA [1.3K]

Answer:

24 hours

Explanation:

The computation is shown below:

The needed mole of O_2 is

= 5 ÷22.4 = n

Also 1 mole of O_2 required four electric charge

Now the charge needed is

= n × 4 × 96,500 C

= 4 × 96,500 × 5 c ÷ 22.4

= 86160.714 C

Now

q = i t

t = q ÷ i

= 86160.714 C ÷ 0.995

= 86593.7 seconds

= 24 hours

Hence, the correct option is A.

3 0
3 years ago
A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the
Kamila [148]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final temperature of the gas is 740 K</em></u>

Explanation:

Given:

Initial pressure of the gas (P_{1}) = 1.8 atm

Final pressure of the gas (P_{2})  = 4 atm

Initial temperature of the gas (T_{1}) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas (T_{2})

We know;

From the ideal gas equation;

we know;

P  × V = n × R × T

So;

we can tell from the above equation;

 <u>   P ∝ T</u>

(i.e.)

      <em> </em>\frac{P}{T}<em> = constant</em>

        \frac{P_{1} }{P_{2} } = \frac{T_{1} }{T_{2} }

Where;

P_{1}  = initial pressure of a gas

P_{2} = final pressure of a gas

T_{1} = initial temperature of a gas

T_{2} = final temperature of a gas

        \frac{1.8}{4} = \frac{333}{T_{2} }

   T_{2} =\frac{333*4}{1.8}

    T_{2} = 740 K

<u><em>Therefore the final temperature of the gas is 740 K</em></u>

8 0
3 years ago
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