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2 years ago

Is kelp a living or non-living thing?

Chemistry

1 answer:

Fed [463]2 years ago

3 0

Answer:

non living

Explanation:

they trying to trick you

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Help me pls

harkovskaia [24]

Answer:

this is answers to the second picture

Explanation:

1- energy

2- structure

3- reaction

4- substance

5- reactants

6- products

7- changed

8- At first there is no reaction because of the protective oxide coating on the aluminum. But soon an exothermic reaction between bromine and aluminum causes brown fumes of bromine to be given off. But as a new substance, aluminum bromide is produced. White smoke is produced.

9- heated to 120 degrees Celsius it melts and first forms a pale-yellow liquid.

In this form the atoms are covalently bonded, and the liquid can flow easily. However, heated up beyond 150 degrees Celsius, the result is a dark red

10- When a chemical change occurs it is more difficult to revert the substance back to its original state. (To be honest I don't know)

3 0

2 years ago

Make the following conversion. 120 mL = _____ cm³ 12.0 1200 120 1.20

Lina20 [59]

The correct answer would be 120.

4 0

3 years ago

Read 2 more answers

This electron cloud model of aluminum has _______ energy levels with _______ electrons on its outer energy level. A) 1, 2 B) 2,

Talja [164]

Answer:

Explanation:

Have anice day

7 0

2 years ago

What is the chemical name of the compound K2504?

12345 [234]

Answer:

Potassium sulfate

Explanation:

I searched it up on the internet

Can I please have a brainliest

6 0

2 years ago

N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

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

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

7 0

2 years ago

Read 2 more answers