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ivann1987 [24]
3 years ago
9

How many molecules of Acetic Acid (CH3COOH) are in a sample that has a mass of 172.90 g?

Chemistry
1 answer:
STatiana [176]3 years ago
3 0

Answer:

17.46× 10²³ molecules

Explanation:

Given data:

Mass of acetic acid = 172.90 g

Number of molecules = ?

Solution:

First of all we will calculate the number of moles of acetic acid :

Number of moles = mass / molar mass

Number of moles = 172.90 g/ 60.1 g/ol

Number of moles = 2.9 mol

Now the given problem will solve by using Avogadro number.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ molecules

2.9 mol × 6.022 × 10²³ molecules  / 1 mol

17.46× 10²³ molecules

Avogadro number:

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

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How many moles of HCl are in 30.00mL of a 0.1000M HCl solution? A. 0.003000mol. B. 300.0mol C. 0.03000mol D. 3.000mol
Katyanochek1 [597]

Given the volume of HCl solution = 30.00 mL

Molarity of HCl solution = 0.1000 M

Molarity, moles and volume are related by the equation:

Molarity = \frac{Moles of solute}{Volume of solution (L)}

Converting volume of HCl from mL to L:

30.00 mL * \frac{1 L}{1000mL}=0.030000 L

Calculating moles of HCl from volume in L and molarity:

0.03000 L * \frac{0.1000mol}{L}= 0.003000 mol HCl

The final moles would be reported to 4 sig figs. So the correct answer will be 0.03000 mol HCl

Correct option: C. 0.03000mol

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3 years ago
To prevent poisoning, what three items should NEVER be taken into the lab?
Roman55 [17]

The answer of this answer is given in the attached file

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7 0
3 years ago
What metalloid has commonly been used as an insecticide due<br> to its effectiveness as a poison.
Taya2010 [7]

Answer:

Arsenic.

Explanation:

Hello there!

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In addition, high levels of arsenic in food could cause arsenic poisoning in humans as well, that is why such practice must be properly performed and by using the correct security protocol.

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5 0
3 years ago
An aqueous solution of a soluble compound (a nonelectrolyte) is prepared by dissolving 33.2 g of the compound in sufficient wate
stiv31 [10]

Answer:

2710.2g/mol

Explanation:

Step 1:

Data obtained from the question. This include the following:

van 't Hoff factor (i) = 1 (since the compound is non-electrolyte)

Mass of compound = 33.2g

Volume = 250mL

Osmotic pressure (Π) = 1.2 atm

Temperature (T) = 25ºC = 25ºC + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Molar mass of compound =.?

Step 2:

Determination of the molarity of the compound.

The molarity, M of the compound can be obtained as follow:

Π = iMRT

1.2 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 1.2 / (0.0821 x 298)

M = 0.049mol/L

Step 3:

Determination of the number of mole of compound in the solution. This can be obtain as follow:

Molarity = 0.049mol/L

Volume = 250mL = 250/1000 = 0.25L

Mole of compound =..?

Molarity = mole /Volume

0.049 = mole / 0.25

Cross multiply

Mole = 0.049 x 0.25

Mole of compound = 0.01225 mole.

Step 4:

Determination of the molar mass of the compound. This is illustrated below:

Mole of the compound = 0.01225 mole.

Mass of the compound = 33.2g

Molar mass of the compound =.?

Mole = Mass /Molar Mass

0.01225 = 33.2/Molar Mass

Cross multiply

0.01225 x molar mass = 33.2

Divide both side by 0.01225

Molar mass = 33.2/0.01225

Molar mass of the compound = 2710.2g/mol

Therefore, the molar mass of the compound is 2710.2g/mol

6 0
3 years ago
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction N 2 ( g
GuDViN [60]

Answer:

ΔG = -52.9 kJ/mol

Explanation:

Step 1: Data given

Temperature = 298 K

All species have a partial pressure of 1 atm

Δ G ° = − 69.0 kJ/mol

Step 2: The balanced equation

N2(g) + 3H2(g) ⇆ 2NH3 (g)

Step 3: Calculate Q

we will use the expression: ΔG = ΔG° + RT*ln(Q)

⇒with Q = the reaction coordinate: Q = (PNH3)²/ ((PN2)*(Ph2)³) = 666.67

Step 4: Calculate ΔG

So, ΔG = -69.0 kJ/mol + (0.008314 kJ/mol*K)*(298 K)*ln(666.67) = -52.9 kJ/mol

(R = the gas constant = 8.314 J/mol* K OR 0.008314 kJ/mol*K)

4 0
3 years ago
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