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Colt1911 [192]
3 years ago
12

Which compound(s) contain the most hydrogen atoms per molecule or formula unit? hydrogen selenide ammonium bromide strontium dih

ydrogen phosphate sodium bicarbonate?
Chemistry
1 answer:
igomit [66]3 years ago
3 0

The answer is strontium dihydrogen phosphate

That is strontium dihydrogen phosphate, is the compound containing the most hydrogen atoms per molecule or formula unit.

The formulas of the following compounds are as follows:

Hydrogen selenide H₂Se

Ammonium bromide NH₃Br  

Strontium dihydrogen phosphate  Sr(H₂PO4)₂

Sodium bicarbonate  NaHCO₃

As it can be seen from the formula of strontium dihydrogen phosphate  Sr(H₂PO4)₂ , that it contains the most hydrogen atoms per molecule or formula unit.


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PLEASE HELP MEEE I WILL MARK BRAINLIST FOR THIS ITS SUPER SIMPLE....BUT U MUST TYPE EVERYTHING
Alina [70]

Answer:

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Explanation:

7 0
3 years ago
Read 2 more answers
7 g of table salt is add to water in a dish. The dish is then heated until all the water evaporates. Assuming no chemical reacti
Aleks04 [339]

Answer:

7 grams

Explanation:

The water evaporates leaving the same amount of salt behind

hope this helps...

8 0
3 years ago
A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo
atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

             Weight = 0.09 \times 16.04 + 0.91 \times 29

                          = 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

                    \frac{800}{27.8336}

                     = 28.74 kmol/hr

Now, applying component balance as follows.

                0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}

                   F_{p} = 43.11 kmol/hr

                   F_{A} + F_{B} = F_{p}

                          F_{B} = 43.11 - 28.74

                                      = 14.37 kmol/hr

So, mass flow rate of pure (B), is F_{B} = 14.37 \times 29

                                                    = 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

             Molecular weight of product stream = Mol. weight \times 43.11 kmol/hr

                                  = 0.06 \times 16.04 + 0.94 \times 29

                                  = 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of O_{2} into the product stream is as follows.

                    0.21 \times 0.94 \times 43.11

                      = 8.5099 kmol/hr \times 329 g/mol

                      = 272.31 kg/hr

Therefore, calculate the mass % of O_{2} into the stream as follows.

                 \frac{272.31}{1216.67} \times 100

                     = 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of O_{2} in the product gas is 22.38%.

4 0
3 years ago
What is the volume of a sample that has a mass of 20 grams and density of 4 g/mL?
balandron [24]

Answer:

<h2>5 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question we have

volume =  \frac{20}{4}  \\

We have the final answer as

<h3>5 mL</h3>

Hope this helps you

7 0
3 years ago
The atoms in non metals tend to _____ electrons. Is it gain?
bekas [8.4K]
YES! good job
they do tend to gain electrons!

4 0
4 years ago
Read 2 more answers
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