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2 years ago

You have 47.0 mL of a 2.00 M concentrated or "stock" solution that must be diluted to 0.500 M. How much water should you add?

Chemistry

1 answer:

Inessa [10]2 years ago

6 0

The required volume of water to make the dilute solution of 0.5 M is 188 mL.

<h3>How do we calculate the required volume?</h3>

Required volume of water to dilute the stock solution will be calculated by using the below equation as:

M₁V₁ = M₂V₂, where

M₁ & V₁ are the molarity and volume of stock solution.
M₂ & V₂ are the molarity and volume of dilute solution.

On putting values from the question to the above equation, we get

V₂ = (2)(47) / (0.5) = 188mL

Hence required volume of water is 188 mL.

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B 18 is the answer ......

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3 years ago

What volume, in mL, of carbon dioxide gas is produced at STP by the decomposition of 0.242 g calcium carbonate (the products are

damaskus [11]

Answer:

54.21 mL.

Explanation:

We'll begin by calculating the number of mole in 0.242 g calcium carbonate, CaCO3.

This is illustrated below:

Mass of CaCO3 = 0.242 g

Molar mass of CaCO3 = 40 + 12 +(16x3) = 40+ 12 + 48 = 100 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 0.242/100

Mole of CaCO3 = 2.42×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

CaCO3 —> CaO + CO2

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole CaO and 1 mole of CO2.

Next, we shall determine the number of mole of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole of CO2.

Therefore,

2.42×10¯³ mole of CaCO3 will also decompose to produce 2.42×10¯³ mole of CO2.

Therefore, 2.42×10¯³ mole of CO2 were obtained from the reaction.

Finally, we shall determine volume occupied by 2.42×10¯³ mole of CO2.

This can be obtained as follow:

1 mole of CO2 occupies 22400 mL at STP.

Therefore, 2.42×10¯³ mole of CO2 will occupy = 2.42×10¯³ x 22400 = 54.21 mL

Therefore, 54.21 mL of CO2 were obtained from the reaction.

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3 years ago

How many moles are in 29.5 grams of Ax?

JulsSmile [24]

The number of moles present in 29.5 grams of argon is 0.74 mole.

The atomic mass of argon is given as;

Ar = 39.95 g/mole

The number of moles present in 29.5 grams of argon is calculated as follows;

39.95 g ------------------------------- 1 mole

29.5 g ------------------------------ ?

$= \frac{29.5}{39.95} \\\\= 0.74 \ mole$

Thus, the number of moles present in 29.5 grams of argon is 0.74 mole.

<em>"Your question seems to be missing the correct symbol for the element" </em>

Argon = Ar

Learn more here:brainly.com/question/4628363

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3 years ago

A balloon is filled with 35.0 L of helium in the morning when the temperature is 35.00 oC.

nexus9112 [7]

Answer: V = 33.9 L

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3 years ago

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