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elena-14-01-66 [18.8K]
3 years ago
11

How many atoms are in 4Ca3(PO4)2

Chemistry
1 answer:
Allisa [31]3 years ago
4 0

Answer:

52

I personally don't know how to do this but this is what I was told the answer is

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What caused the falling of the twin towers?
Sergeu [11.5K]

Answer:

ok horita te ayudo ok esperame okey bueno

4 0
3 years ago
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Susan was frantic: the small diamond had disappeared from her wedding ring! It had been there this morning, before she made brea
Bezzdna [24]

Answer:

No. 1 Susan was <u>Frantic</u>

No.2 panicked

No.3 determined

No.4 Despair

No.5 Releif

I really hope this helps, Good luck! :D

7 0
3 years ago
A distinct molecular species that is formed in one step of a reaction mechanism and then reacts in a subsequent step is called
Flura [38]

Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates.

I hope that helps

ps. not my words, my older cousins

8 0
3 years ago
HELP I WILL MARK YOU AS THE BRAINLIEST ANSWER
borishaifa [10]

Hey there! Here is the answer:

B.) decrease; increase

Here's why:

As you move from left to right across a period on the periodic table the size of an atom will decrease. As you move from top to bottom down a group on the periodic table the size of an atom will increase.

I really hope this answer helps you out! :)

8 0
3 years ago
how many grams of ammonium carbonate are needed to decompose in order to produce 6.52 g of carbon dioxide?
defon

Answer:

<u>= 14.24g of </u>(NH4)_{2}CO_{3}<u> is required.</u>

Explanation:

Reaction equation:

(NH4)_{2}CO_{3} → 2NH_{3} + CO_{2} + H_{2} O

Mole ratio of ammonium carbonate to carbon dioxide is 1:1

1 mole of CO2 - 44g

?? mole of CO2 - 6.52g

= 6.52/44 = 0.148 moles was produced from this experiment.

Therefore, if 1 mole of (NH4)_{2}CO_{3} - 96.09 g

0.148 mol of (NH4)_{2}CO_{3} --  ?? g

=0.148 × 96.09

<u>= 14.24g of </u>(NH4)_{2}CO_{3}<u> is required.</u>

5 0
3 years ago
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