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melamori03 [73]

3 years ago

9

A green powder was placed into a beaker of water. After it was stirred

1 answer:

bekas [8.4K]3 years ago

7 0

Answer:

A solution was NOT formed.

Explanation:

For a solution to be formed, the green powder would need to dissolve in the water to form a homogenous mixture. But since the water was cloudy and had lumps of the green powder, it means that the powder did not dissolve, but instead remained insoluble.

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How many moles are in 3.46 g of chromium?

Bezzdna [24]

Answer:

0.06654345229738384 moles of chromium.

6 0

3 years ago

A student dissolves 0.0688 mol of sodium hydroxide in water to make a 0.250 L solution. What is the molarity of the solution?

den301095 [7]

Number of moles = 0.0688 moles of NaoH

volume = 0.250 L

Molarity = moles of solute / volume ( L )

M = 0.0688 / 0.250

M = 0.28 M

Answer B

7 0

3 years ago

How many liters of CO2 are in 4.76 moles? (at STP)

dem82 [27]

<u>Answer:</u> The volume of carbon dioxide gas at STP for given amount is 106.624 L

<u>Explanation:</u>

We are given:

Moles of carbon dioxide = 4.76 moles

<u>At STP:</u>

1 mole of a gas occupies a volume of 22.4 Liters

So, for 4.76 moles of carbon dioxide gas will occupy a volume of = $\frac{22.4L}{1mol\times 4.76mol=106.624L$

Hence, the volume of carbon dioxide gas at STP for given amount is 106.624 L

6 0

3 years ago

Which property is NOT used to separate a mixture?

NeTakaya

Answer: ability to undergo a chemical reaction

Explanation:

6 0

3 years ago

Be sure to answer all parts. Calculate the pH of the following two buffer solutions: (a) 1.4 M CH3COONa/1.6 M CH3COOH. (b) 0.1 M

aalyn [17]

Answer:

a) pH = 4.68 (more effective)

b) pH =4.44.

Explanation:

The pH of buffer solution is obtained by Henderson Hassalbalch's equation.

The equation is:

$pH =pKa +log\frac{[salt]}{[acid]}$

a) pKa of acetic acid = 4.74

[salt] = [CH₃COONa] = 1.4 M

[acid] = [CH₃COOH] = 1.6 M

$pH = 4.74 + log \frac{1.4}{1.6}= 4.68$

This is more effective as there is very less difference in the concentration of salt and acid.

b) pKa of acetic acid = 4.74

[salt] = [CH₃COONa] = 0.1 M

[acid] = [CH₃COOH] = 0.2 M

$pH = 4.74 + log \frac{0.1}{0.2}= 4.44$

3 0

3 years ago