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3 years ago

Can we use electrical equipment near water? Explain Why?

Chemistry

1 answer:

tiny-mole [99]3 years ago

5 0

Answer:

You may, but it is too risky.

Even though you are being cautious around using electric equipment around water, you'll never know what can happen. You might accidentally drop that piece of electrical equipment you are using into the water. Water can be splashed around by someone or something without you noticing it and it may affect the object you are using. Sometimes, if water comes in contact with an electrical object, it may cause you electric shocks or the equipment you are using has a chance of exploding and may hurt you. You can guarantee that waterproof electrical equipment is safe to use, but it is better not to risk it too much.

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Draw the Lewis Dot structure for the following diprotic acid: HOOC-(CH2)4-COOH. 35.50 mL of this acid is titrated to end point w

Sliva [168]

hey there!:

Attached resolution.

5 0

3 years ago

Calculate the pH of a 3.58x10^-9 M Nitric acid (a strong acid) solution

lubasha [3.4K]

Answer:

pH= 8.45

Explanation:

when working with strong accids pH = -log(Concentration)

so -log(3.58e-9) = 8.446

7 0

3 years ago

1.674×10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 86.6 s. Under identical conditions, 1.715×10-

siniylev [52]

<h2>Answer:</h2>

44.06 g/mol

<h3>Explanation:</h3>

We are given;

Number of moles of unidentified gas as 1.674×10^-4 mol
Time of effusion of unidentified gas 86.6 s
Number of moles of Argon gas as 1.715×10^-4 mol
Time of effusion of Argon gas is 84.5 s

We are supposed to calculate the molar mass of unidentified gas

<h3>Step 1: Calculate the effusion rates of each gas</h3>

Effusion rate = Number of moles/time

Effusion rate of unidentified gas (R₁)

= 1.674×10^-4 mol ÷ 86.6 s

= 1.933 × 10^-6 mol/s

Effusion rate of Argon gas (R₂)

= 1.715×10^-4 mol ÷ 84.5 sec

= 2.030 × 10^-6 mol/s

<h3>Step 2: Calculate the molar mass of unidentified gas</h3>

Assuming the molar mass of unidentified gas is x;
We can use the Graham's law of effusion to find x;
According to Graham's law of diffusion;

$\frac{R_{1}}{R_{2}}}=\frac{\sqrt{MM_{Ar}}}{\sqrt{X}}$

But, Molar mass of Argon is 39.948 g/mol

Therefore;

$\frac{1.933*10^-6mol/s}{2.030*10^-6mol/s}}=\frac{\sqrt{39.948}}{\sqrt{X}}$

$0.9522=\frac{\sqrt{39.948}}{\sqrt{X}}$

Solving for X

x = 44.06 g/mol

Therefore, the molar mass of the identified gas is 44.06 g/mol

3 0

4 years ago

How are the atoms of carbon different from the atoms of nitrogen

mars1129 [50]

<h2>Answer:</h2>

<em>Per my research, it was said that each of the atoms of the carbon has over 6 protons, six neutrons, and 6 protons. Each atom of nitrogen has 7 protons, 7 electrons, and 7 neutrons. The atomic mass of carbon and nitrogen is 12 and 14 respectively. As the number of electrons in the atoms of the two elements is different they have different chemical properties.</em>

<h2>Explanation:</h2>

<em>As you may know, I did my research and this is what it gave me. I would suppose it is correct since it explains that the carob has over 6 protons and so on. Correct me if I am incorrect and I hope this helped. Have a good one!</em>

8 0

3 years ago

The density of sea water is 1.029 kilograms per liter. What is the density of

Lelu [443]

1029 grams per kilometer

5 0

3 years ago