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Ksenya-84 [330]

3 years ago

6

A student compares the booking point of substances having different intermolecular forces. Which independent variable did the st

udent most likely use

1 answer:

Paha777 [63]3 years ago

4 0

Answer:

a certain booking point

Explanation:

i need brainly

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A 5.098 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give

GarryVolchara [31]

<u>Answer:</u> The molarity of solution is 0.954 M

<u>Explanation:</u>

We are given:

5.098 mass % solution of potassium hydroxide

This means that 5.098 grams of potassium hydroxide is present in 100 grams of solution

To calculate volume of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.05 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

$1.05g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.05g/mL}=95.24mL$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of potassium hydroxide = 5.098 g

Molar mass of potassium hydroxide = 56.1 g/mol

Volume of solution = 95.24 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{5.098\times 1000}{56.1\times 95.24}\\\\\text{Molarity of solution}=0.954M$

Hence, the molarity of solution is 0.954 M

7 0

4 years ago

. How many grams of H2O can you make with 120g of O2.

Sladkaya [172]

Answer:

i don't thing i understand the question.

Explanation:

7 0

3 years ago

How many grams of H2O are there in 6.02 x 1023 molecules of water?

tamaranim1 [39]

Answer:

By definition, there are 6.022×1023 such molecules, or NA such molecules in ONE mole of water. And thus in such a quantity there are NA oxygen atoms, and 2×NA hydrogen atoms...and the mass associated with this numerical quantity of water molecules is approx. 18⋅g ...

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3 years ago

If you have 3 moles of calcium carbonate, how many grams of calcium bicarbonate are formed?

Flura [38]

Answer:

hhjcioz xlioyudiyyxyisrupautwtritu regards Roy

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3 years ago

You are asked to prepare 500 mL 0.200 M acetate buffer at pH 5.00 using only pure acetic acid ( MW=60.05 g/mol, pKa=4.76), 3.00

Vilka [71]

Answer:

You need to weight 6,005 g of acetic acid

Explanation:

Using Henderson-Hasselbalch formula you will obtain:

5,00 = 4,76 +log₁₀ $\frac{[Ac^-]}{[Acac]}$

<em>Where Ac⁻ is the salt of acetic acid (Acac).</em>

Solving:

1,738 = $\frac{[Ac^-]}{[Acac]}$ <em>(1)</em>

Also, yo know that:

0,200 M = [Ac⁻] + [Acac] <em>(2)</em>

Replacing (2) in (1):

[Acac] = 0,0730 M.

Thus:

[Ac⁻] = 0,127 M

The moles of each compound are:

Acac = 0,0730 M × 0,500 L = <em>0,0365 mol</em>

Ac⁻ = 0,127 M × 0,500 L = <em>0,0635 mol</em>

To prepare these moles it is necessary to use:

Acac + NaOH → AcNa + H₂O

The initial moles of Acac must be:

0,0365 moles + 0,0635 moles = 0,100 moles

<em>To obtain 0,0635 moles of Ac⁻ you need to take this quantity of NaOH moles.</em>

Thus, to obtain a acetate buffer of 5,00 you need to add 0,100 moles of acetic acid and 0,0635 moles of NaOH because This NaOH will react with acetic acid producing 0,0635 moles of Ac⁻ and surplus 0,0365 moles of acetic acid.

Now, to obtain 0,100 moles of acetic acid from pure acetic acid:

0,100 moles × $\frac{60,05 g}{1 mol}$ = <em>6,005 g</em>

<em>You need to weight 6,005 g of acetic acid</em>

I hope it helps!

7 0

4 years ago