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Mandarinka [93]
3 years ago
12

A 0.35 g sample of Li(s) is placed in an Erlenmeyer flask containing 100 mL of water at 25°C. A balloon is placed over the mouth

of the flask to collect the hydrogen gas that is generated
Chemistry
1 answer:
Harrizon [31]3 years ago
7 0

Answer:

C- Using a 0.35 g sample of Li(s) cut into smaller pieces will increase the rate of reaction

C- Equal quantities of gas will be produced with potassium K, used instead of Lithium, Li.

Explanation:

<u>First part</u>

When a metal is present in small pieces, the rate of the chemical reaction increases. This is due to small pieces provide more surface area for reaction than huge chunks of Li.

moles of reactant decreases product formation rate also decreases and temperature decreases rate also decreases so remaining option are wrong

<u>Second part </u>

2Li + 2H2O --------------> 2LiOH + H2

2K + 2H2O ---------------> 2KOH + H2  

Hence, both gives same quantity of gas

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A compound is found to be 63.6 % N and 36.4% O. What is the empirical formula?
FromTheMoon [43]

Answer:

NO2

Explanation:

Calculate moles of Nitrogen and oxygen (assume total mass is 100g as you only have percentages therefore there is 63.6g of N and 36.4g of O) This is the ratio that they are in but you need the ratio to be in whole numbers therefore divide each number of moles by 2.275 (the moles of oxygen) to calculate the empirical formula. Hope this helps!

8 0
3 years ago
Can someone tell me a summary of volcanoes and plate tectonics? (Please quickly!)
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TECTONIC PLATES

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VALCONES AND TECTONIC

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3 years ago
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What are the terms for A, B, C?<br> C<br> А<br> В
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6 0
3 years ago
If Carl buys a 946 ml bottle of rubbing alcohol, how much of the aqueous solution is water?
Darina [25.2K]

The question is incomplete, here is the complete question:

A bottle of rubbing alcohol having aqueous solution of alcohol contains 70% (v/v) alcohol. If Carl buys a 946 ml bottle of rubbing alcohol, how much of the aqueous solution is water?

<u>Answer:</u> The amount of water present in the given bottle of rubbing alcohol is 283.8 mL

<u>Explanation:</u>

We are given:

Volume of bottle of rubbing alcohol = 946 mL

70% (v/v) alcohol solution

This means that 70 mL of rubbing alcohol is present in 100 mL of solution

Amount of water present in solution = [100 - 70] = 30 mL

Applying unitary method:

In 100 mL of solution, the amount of water present is 30 mL

So, in 946 mL of solution, the amount of water present will be = \frac{30}{100}\times 946=283.8mL

Hence, the amount of water present in the given bottle of rubbing alcohol is 283.8 mL

4 0
3 years ago
A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

5 0
3 years ago
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