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2 years ago

Which is an example of a mixture made of parts that can easily be separated?

2 answers:

7 0

Hi there.

The imagine trying to separate flour, eggs, & sugar!

The <em>easiest</em> option here is D. Snack mix.

You could just pick out all the ingredients. (maybe eat a couple pretzels..)

~

RideAnS [48]2 years ago

3 0

D Snack Mix because they are individual parts that can easily be seperated

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In the equation N2(g) + 3H2(g)--- 2NH3(g), how many liters of hydrogen

Delicious77 [7]

Answer:

30 L H2

Explanation:

10 L N2 x <u>3 L H2</u> = 30 L H2

. 1 L N2

Try to verify my answer, Stoichiometry is not easy for me.

3 0

2 years ago

Read 2 more answers

How many liters would you need to make a 0.8 M solution with 20 grams of lithium chloride?

worty [1.4K]

Answer:

0.5875L

Explanation:

concentration = mole/ volume

n(LiCl) = 20 / (7 + 35.5) = 0.47 mol

volume = mole / conc.

volume = 0.47 /0.8

= 0.5875 dm³ = 0.5875L

5 0

2 years ago

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Minchanka [31]

Answer:

C. The third option

Explanation:

7 0

3 years ago

What was the eutectic temperature (temperature from the two lines of best fit cross) for the mixture

Ymorist [56]

Answer:

hello your question is incomplete below is the missing part of the question

answer : 104°c

Explanation:

The Eutectic temperature for the mixture is 104°c

From the chart attached below it can be seen that the temperature from the two lines of best fit cross is 104°c

7 0

2 years ago

The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon

tigry1 [53]

Answer:

$\boxed{\text{300 g}}$

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r: 32 60

CH₃OH + CO ⟶ CH₃COOH

m/g: 160

(a) Moles of CH₃OH

$\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}$

(b) Moles of CH₃COOH

$\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}$

(c) Mass of CH₃COOH

$\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}$

3 0

3 years ago