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3 years ago

Why is deuterium oxide called “heavy water”?

Chemistry

2 answers:

kompoz [17]3 years ago

6 0

Answer:

it is going to B because the hydrogen and some antonyms are heavier than others and that makes it heavy water basically

Explanation:

hope that helps I also looked it up and I've also been learning the same thing so I have my notes that I can study

konstantin123 [22]3 years ago

3 0

Answer:

a) su as tomos de oxigeno son mas pesdos que lo otros

Explanation:

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dmitriy555 [2]

Answer:

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Explanation:

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5 0

3 years ago

Balance the following chemical equations

jok3333 [9.3K]

Answer:

2N2 + O2 ⇒ 2N2O

Explanation:

Since I cannot see the product's of the second chemical equation I will only solve the first one.

In order to balance a chemical equation you need to make sure that the atoms on both sides are equal.

In this case we have...

N2 + O2 = N2O

N = 2

O = 2

N = 2

O = 2

Change the coefficient:

N2O = 2N20

2 × 2 = 4

1 × 2 = 2

2N2 + O2 ⇒ 2N2O

Hope this helps.

4 0

2 years ago

There are many methods used to obtain energy for use in our our daily lives. Some of these methods have possible environmental c

Len [333]

(A)wind and solar should be the answer

8 0

3 years ago

Liquids: A) must be contained to stay in one spot B) have a variable volume, but a fixed mass C) are usually more dense than sol

Margaret [11]

B). Have a variable volume, meaning they fill up the area they are in.

8 0

3 years ago

Read 2 more answers

quien y en que cantidad sera el reactivo limite, si utilizamos 125 g de ácido, H (CH3COO) y 275 g de hidróxido , Al(OH)3

Deffense [45]

Answer:

The limiting reactant is acetic acid. All 125 g will react.

Explanation:

1. Assemble the information

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

Mᵣ: 60.05 78.00

3CH₃COO-H + Al(OH)₃ ⟶ (CH₃COO)₃Al + 3H₂O

Mass/g: 125 275

2. Calculate the moles of each reactant

$\text{Moles of CH$_{3}$COOH} = \text{125 g CH$_{3}$COOH} \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{60.05 g CH$_{3}$COOH}} = \text{2.082 mol CH$_{3}$COOH }\\\\\text{Moles of Al(OH)}_{3} = \text{275 g Al(OH)}_{3} \times \dfrac{\text{1 mol Al(OH)}_{3}}{\text{78.00 g Al(OH)}_{3}} = \text{3.526 mol Al(OH)}_{3}$

3. Calculate the moles of (CH₃COO)₃Al from each reactant

$\textbf{From CH$_{3}$COOH:}\\\text{Moles of (CH$_{3}$COO)$_{3}$Al} = \text{2.082 mol CH$_{3}$COOH} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al}}{\text{3 mol CH$_{3}$COOH}}\\\\= \text{0.6939 mol (CH$_{3}$COO)$_{3}$)Al}\\\textbf{From Al(OH)}_{3}:\\\text{Moles of (CH$_{3}$COO)$_{3}$Al } =\text{3.526 Al(OH)}_{3} \times \dfrac{\text{1 mol (CH$_{3}$COO)$_{3}$Al }}{\text{1 mol Al(OH)}_{3}}\\\\= \text{3.526 mol (CH$_{3}$COO)$_{3}$Al}$

$\text{Acetic acid is the $\textbf{limiting reactant}$ because it gives fewer moles of} \\\text{(CH$_{3}$COO)$_{3}$Al. All $\textbf{125 g}$ will react.}$

5 0

3 years ago