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3 years ago

BRAINLIEST IF ANSWERED IN THE NEXT 5M

Chemistry

1 answer:

rosijanka [135]3 years ago

4 0

(NH₄)₃PO₄ → 3NH₄⁺ + PO₄³⁻
k=3 k=1

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How many elements are in Li2SO4?how many elements are in li2s 04 how many elements are in li2s 04

Free_Kalibri [48]

Answer:

Explanation:

the three elements involved in this compound are Li, S, O.

lithium, sulfur, and oxygen. Which create the ionic compound, "Lithium Sulfate."

7 atoms total, since there are two lithium, four oxygen, and one sulfate atom. this is a white inorganic salt.

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3 years ago

Write the formula for a compound of manganese (IV) hydroxide.?

Oxana [17]

You first add the manganese and exchange the number of electrons needed with the hydroxide. While the hydroxide needs only 1 electron the manganese needs 4, so after you exchange the electrons the manganese will be just 1 atom while the hydroxide is 4. Mn(OH)4

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3 years ago

What is the process of breaking down food to release energy by adding oxygen

a_sh-v [17]

I believe this process is called cellular respiration.

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3 years ago

How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

Gnoma [55]

Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

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3 years ago

U-235 undergoes many different fission reactions. For one such reaction, when U-235 is struck with a neutron, Ce-144 and Sr-90 a

Lostsunrise [7]

Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

The given reaction is $^{235}_{92}U + ^{1}_{0}n \rightarrow ^{144}_{58}Ce + ^{90}_{38}Sr + x^{1}_{0}n + y^{0}_{-1}\beta$

Now, we balance the mass on both reactant and product side as follows.

235 + 1 = $144 + 90 + (x \times 1) + (y \times 0)$

236 = 234 + x

x = 236 -234

= 2

So, now we balance the charge on both reactant and product side as follows.

92 + 0 = $58 + 38 + (x \times 0) + (y \times -1)$

92 = 96 - y

y = 4

Thus, we can conclude that there are 2 neutrons and 4 beta-particles are produced in the given reaction.

Therefore, reaction equation will be as follows.

$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{144}_{58}Ce + ^{90}_{38}Sr + 2^{1}_{0}n + 4^{0}_{-1}\beta$

8 0

3 years ago