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3 years ago

Is changing the color of the coin from copper to silver a physical or chemical change?

Chemistry

1 answer:

melamori03 [73]3 years ago

7 0

Answer:

physical

Explanation:

The zinc (which was already silver) coated the penny. The NaOH dissolved the zinc. (that was what made the water murky) And small pieces of zinc adhered to the penny (coated) It was a physical change because no NEW color was created.

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The only type of movement particles in a solid do is

Nina [5.8K]

Answer

:small vibrational movements.

Explanation:

The particles of a solid are not able to move out of their positions relative to one another, but do have small vibrational movements.

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3 years ago

Which of the following best describes an atom?

Tanzania [10]

Answer: The smallest unit of matter.

Explanation:

7 0

4 years ago

Plz HELP ITS SCIENCE!

Alexxx [7]

Cell wall:
-acts like a skin
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6 0

3 years ago

Read 2 more answers

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago

a student determined in the laboratory that the percent by mass of water in cus04 5h20 is 38.0% if the accepted value is 40% wha

IrinaVladis [17]

Percent error is [(abs(experimental - actual))/(actual)] * 100
(38.0 - 40.0 ) / 40.0 = 2.0 / 40 = 0.05
0.05 * 100 = 5
since there are 2 sigfigs, the answer is 5.0 %

I made the 2 positive since the absolute value of 38.0 - 40.0 was taken
There are only 2 sigfigs because of the subtraction of 38.0 - 40.0 gives 2.0, only 2 sigfigs

4 0

4 years ago