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3 years ago

Chem question!! please help!!!

Chemistry

1 answer:

podryga [215]3 years ago

6 0

Cu is the answer... :)

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IT is called Iron(iii)oxide

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3 years ago

Is this equation balanced or unbalanced?<br> C+02=CO2

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This equation is balanced

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2 years ago

If a person had the values for an objects density and volume, what value can be calculated

zavuch27 [327]

Hello there! With the values of density and volume, you would be able to find the object's mass.

Density is found by dividing the mass by the volume, so you could place in the values of the density and the volume to get the mass.

For example:

500 = mass/10

The 500 being density and 100 being volume. You would use simple math rules and multiply 10 by 500, and you'd get 5000, therefore using the density and volume values and giving you the mass.

I hope I could help you and have a great day!

5 0

2 years ago

Read 2 more answers

Acetylene gas (ethyne; HC = CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapor. The heat of reaction f

Yanka [14]

The mass of CO2 produced by 26g of acetylene is 88g.

Given ,

In an oxyacetylene torch, acetylene gas (ethyne; HCCH) burns to produce carbon dioxide and water vapour.

The acetylene combustion reaction is given by,

H2O + HCCH + 5/2 O=O 2CO2

Heat of reaction for acetylene combustion = 1259kj/mol

CO2 has a molecular mass of 44g/mol.

2 moles of CO2 have a molecular mass of 88g.

On combustion, 1 mole of acetylene yields 2 moles of CO2.

Thus, 26g of acetylene produces 88g of CO2 when burned.

As a result, the mass of carbon dioxide produced by 26g of acetylene is 88g.

Learn more about acetylene here :

brainly.com/question/15346128

#SPJ4

6 0

1 year ago

What mass of Cu(s) is electroplated by running 24.5A of current through a Cu2+(aq)solution for 4.00 h?Express your answer to thr

Masja [62]

Answer: 116 g of copper

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 24.5A

t= time in seconds = 4.00 hr = $4.00\times 3600s=14400s$ (1hr=3600s)

$Q=24.5A\times 14400s=352800C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500C=193000C$ of electricity deposits 63.5 g of copper.

352800 C of electricity deposits = $\frac{63.5}{193000}\times 352800=116g$ of copper.

Thus 116 g of Cu(s) is electroplated by running 24.5A of current

Thus remaining in solution = (0.1-0.003)=0.097moles

8 0

2 years ago