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dedylja [7]
3 years ago
14

The percent composition by mass of an unknown chlorinated hydrocarbon was found to be 37.83% C, 6.35% H, and 55.83% Cl by mass.

What is the empirical formula of this compound?
Chemistry
1 answer:
iogann1982 [59]3 years ago
8 0

Answer:

C2H4Cl

Explanation:

The first step in obtaining the empirical formula is to divide the percent by mass of each element by its relative atomic mass;

C- 37.83/12, H- 6.35/1, Cl- 55.83/35.5

3.15, 6.35, 1.57

Divide through by the lowest ratio

3.15/1.57, 6.35/1.57, 1.57/1.57

2, 4, 1

Hence, empirical formula is C2H4Cl

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Question 8<br> What is the effective nuclear charge for a 2nd row electron in sulfur
miss Akunina [59]

Answer:

The effective nuclear charge for a 2nd row electron in Sulfur is +8

Explanation:

Zeff = Z (# of protons) - S (# of shielded electrons)

Since there are 8 electrons in the first and second rows combined, there are 8 shielding electrons.

The number of protons in Sulfur is 16.

Therefore,

Zeff = 16 - 8

Zeff = 8

(It's been awhile, so I am not 100% sure)

7 0
2 years ago
A patient needs to take her cholesterol medication every morning and night. Her prescription bottle most likely says
mamaluj [8]

Answer:

1+0+1

Explanation:

7 0
3 years ago
What is the pH of this solution?
Vesnalui [34]

Answer:

pH = 11.216.

Explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

NH_3+H_2O\rightleftharpoons NH_4^++OH^-

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M

Which is also:

[OH^-]=1.643x10^{-3}M

Thereafter we can compute the pOH first:

pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784

Finally, the pH turns out:

pH=14-2.784\\\\pH=11.216

Regards!

5 0
3 years ago
How many moles of electrons is required to deposit 5.6g of iron from a solution of iron (2) tetraoxosulphate(6)
yuradex [85]

Answer:

0.20 mol

Explanation:

Let's consider the reduction of iron from an aqueous solution of iron (II).

Fe²⁺ + 2 e⁻ ⇒ Fe

The molar mass of Fe is 55.85 g/mol. The moles corresponding to 5.6 g of Fe are:

5.6 g × 1 mol/55.85 g = 0.10 mol

2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are

0.10 mol Fe × 2 mol e⁻/1 mol Fe = 0.20 mol e⁻

7 0
3 years ago
How does the density of a hershey bar compare to the density of the a half bar?
anastassius [24]

The density would be the same for the whole bar as well as one half of the bar. Density is a identity I believe, by this I mean that it stays the same no matter how little or how much of the same substance you have. Since density = mass / volume, half the bar has half of the weight as well as half of the volume of the whole bar, making the density the same.

For example, a block weighs 10 grams and has a volume of 5 ml. the density would be d = 10/5 or, d = 2g/ml

Half of the block weighs 5 grams and has a volume of 2.5 ml. The density is d = 5/2.5, or, d = 2 g/ml.

See, although there are different amounts of the same substance, their density is the same.

8 0
2 years ago
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