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zmey [24]
3 years ago
5

What is the charge of an electron

Chemistry
1 answer:
Afina-wow [57]3 years ago
7 0
The charge of an electron is 4.80320451 × 10<span>^−10 .</span>
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what happen when carbon dioxide is passed through water for short time and for long time?explain with chemical reaction.​
Delicious77 [7]

Answer:

Expert Verified

Explanation:

For short duration: ... When excess of carbon dioxide gas is passed through lime water then the white precipitate calcium carbonate formed first dissolves due to the formation of a soluble salt calcium hydrogen carbonate (Ca(HCO3)2, and the Solution becomes clear again

6 0
3 years ago
Reactant(s)<br> H<br> 0,<br> Product(s)<br> H,0<br> 3.4g<br> 10g
zalisa [80]

Answer:

H,0

3.4g

Explanation:

H,0

3.4g

8 0
3 years ago
An infectious disease is _____.
e-lub [12.9K]

Answer: caused by organismis

Explanation:

3 0
3 years ago
2) Calculate the percent composition of each element in Mgso,
iogann1982 [59]

Answer:

2)

\% Mg=20.2\%\\\\\% S=26.6\%\\\\\% O=53.2\%

3)

\% Ag=93.1\%\\\\\% O=6.9\%

Explanation:

Hello!

2) In this case, since magnesium sulfate is MgSO₄, we can see how magnesium weights 24.305 g/mol, sulfur 32.06 g/mol and oxygen 64.00 g/mol as there is one atom of magnesium as well as sulfur but four oxygen atoms for a total of g/mol; thus the percent compositions are:

\% Mg=\frac{24.305}{120.36 } *100\%=20.2\%\\\\\% S=\frac{32.06}{120.36 } *100\%=26.6\%\\\\\% O=\frac{64.00}{120.36 } *100\%=53.2\%

3) In this case, although the element seems to contain Ag and O, we infer its molecular formula is Ag₂O; thus, since we have two silver atoms weighing 215.74 g/mol and one oxygen atom weighing 16.00 g/mol for a total of 231.74 g/mol, we obtain the following percent compositions:

\% Ag=\frac{215.74}{231.74} *100\%=93.1\%\\\\\% O=\frac{16.00}{231.74} *100\%=6.9\%

Best regards!

7 0
3 years ago
Serial dilution problem: Six test tubes are placed in a rack. To each tube add 4 mL of saline solution. Now to the first tube ad
ArbitrLikvidat [17]

Answer:

The dilution factor of protein in tube # 4 is 125. Molar concentration is 0.0088 M protein

Explanation:

The dilution factor indicates how many times is more concentrated a main solution in relationship with a diluted solution. In this case, the main solution is in tube #1. For calculating the dilution factor and molar concentration in tube #4 we need the main solution concentration which comes from next equation:

Initial volume * initial concentration = final volume * final concentration

0.5 mL * 10M = 5mL * final concentration

1.1 M = final concentration = main solution concentration

Applying the same equation for remain tubes we have 0.22 M for tube #2, 0.044 M for tube # 4 and 0.0088 for tube # 4.

Dilution factor = Main solution concentration/tube 4 concentration

Dilution factor = 1.1/0.0088 = 125

I hope my answer helps you

3 0
3 years ago
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