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2 years ago

Question 7 (2 points) A mocktail is a flavoured cocktail. O True O False

Chemistry

2 answers:

RoseWind [281]2 years ago

5 0

False a mock tail is a cocktail with the liquor.

Reptile [31]2 years ago

4 0

Answer:

false mocktail is liquor yea

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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

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$

Cross multiply

$0.505 * V_{B} = 20.2 * 2\\0.505 * V_{B} = 40.4$

Divide both side by 0.505

$V_{B} = \frac{40.4}{0.505}\\\\V_{B} = 80 mL$

Finally, we shall convert 80 mL to dm³. This can be obtained as follow:

$1000 mL = 1 dm^{3}\\\\Therefore,\\\\80 mL = \frac{80 mL * 1dm^{3}}{1000 mL}\\\\80 mL = 0.08dm^{3}$

Therefore, the volume of NaOH required is 0.08 dm³

Learn more: brainly.com/question/19053582

3 0

3 years ago

What orbits around the nucleus

Flura [38]

Answer:

Electrons orbit the nucleus.

5 0

3 years ago

Read 2 more answers

if i mix blue gatorade with red gatorade what do i get this is for my science exam right now please answer

Gnesinka [82]

Answer:

purple gatorade

Explanation:

7 0

3 years ago

How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.48 L against a constant pressure

Dahasolnce [82]

Given:
ΔV = 5 - 1.48 L = 3.52 L = 0.00352 m³, th chane in volume
p = 0.818 atm = 0.819*101325 Pa = 82985.2 Pa

By definition, the work done is
$W = \int pdV = (82985.2 \, \frac{N}{m^{2}} )*(0.00352 \, m^{3}) = 292.1 \, J$

Answer: 292.1 J

5 0

3 years ago