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3 years ago

How many moles of HCl are present in 0.70 L of a 0.33 M HCl solution? (molar mass of HCl = 36.46 g/mol)

Chemistry

1 answer:

Vilka [71]3 years ago

5 0

Answer:-

0.231 moles

Explanation:-

From the question we are told that

Volume of HCl solution = 0.70 L

Strength of HCl solution = 0.33 M

Here M stands for molarity. Molarity is the number of moles of the substance present per Litre of solution. So in place of M we can write moles / L

Number of moles of HCl solution = Volume of HCl solution x Strength of HCl solution

= 0.70 L x 0.33 moles / L

= 0.231 moles.

0.231 moles of HCl are present in 0.70 L of a 0.33 M HCl solution

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You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar

djverab [1.8K]

1.04 ⨯ 10 $^{-4}$ M

<h3>Explanation</h3>

<em>A</em> = <em>ε</em> $\cdot l \cdot c$ by the Beer-Lambert law, where

<em>A</em> the absorbance,
$l$ the path length,
<em>ε</em> the molar absorptivity of the solute, and
$c$ concentration of the solution.

<em>A</em> and <em>ε </em>are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$ . The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.

6 0

3 years ago

100 POINTS! Final Honor Activity Question

castortr0y [4]

The change in temperature had the greatest effect at changing the volume of the balloon.

The gas laws are used to describe the parameters that has to do with gases.

Given that;

P1 = 98.5 kPa

T1 = 18oC or 291 K

V1 = 74.0 dm3

P2 = 7.0 kPa

V2 = ?

T2 = 18oC or 291 K

P1V1/T1 = P2V2/T2

P1V1T2 =P2V2T1

V2= P1V1T2/P2T1

V2 = 98.5 kPa * 74.0 dm3 * 291 K/ 7.0 kPa * 291 K

V2 = 1041.3 dm3

When;

V1 = 1041.3 dm3

T1 = 291 K

V2 = ?

T2 = 80oC or 353 K

V1/T1 = V2/T2

V1T2 = V2T1

V2 = V1T2/T1

V2 = 1041.3 dm3 * 353 K/291 K

V2 = 1263 dm3

The change in temperature had the greatest effect at changing the volume of the balloon.

Given that

V1 = 100 cm^3

T1 = 273 K

P1 = 1.01 * 10^5 Pa

V2 = ?

P2 = 3.00 x 10^-4 Pa

T2 = -180oC or 255 K

V2= P1V1T2/P2T1

V2 = 1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K

V2 = 3.14 * 10^10 cm^3

Learn more about gas laws:brainly.com/question/12669509

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7 0

2 years ago

Plzz help me out guys ......10 points rewarded.....!!!!!

GaryK [48]

number 5 is 1 : 1

number1 is also 1:1 ibelieve but it could be 2:1 just like number 5 but im postive its 1:1 for nummer 1 and number 5

4 0

3 years ago

Write the number of sig. fig. in four numbers given in the sentence below. An (one) octopus has 8 legs. 13 octopi have 104 legs.

Marrrta [24]

Answer:

1, 1, 2, 3

Explanation:

The numbers 1 and 8 both have 1 sig. fig.

The number 13 has 2 sig. figs.

The number 104 has 3 sig. figs.

6 0

2 years ago

How is density different from weight and mass?

defon

Hello there!

How is density different from weight and mass you say? It's simple!

1. Unlike weight, density is composed of mass and volume.

2. By dividing the mass over the volume, you can determine an object's density, while weight is the product of mass and gravity.

Hope this helps :) if so, can I get Brainliest?

5 0

1 year ago

Read 2 more answers