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Dovator [93]
2 years ago
14

_NH, +_02 - _ NO, +_ H,0

Chemistry
1 answer:
adoni [48]2 years ago
7 0

Answer:

This equation is already balanced.

Explanation:

If your teacher requires that you put coefficients in front of each compound, then put a 1 in front of every compound.

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Calculate the mass in grams of iodine (i 2 ) that will react completely with 20.4 g of aluminum (al) to form aluminum iodide (al
garik1379 [7]
3I₂ + 2Al → 2AlI₃

m(I₂)=3M(I₂)m(Al)/{2M(Al)}

m(I₂)=3*253.8*20.4/{2*27.0}=287.64 g
8 0
3 years ago
Help please all please
Luden [163]

Answer:

I don’t want to download a pdf that I don’t know what it is…

Also, brainly strictly says that we can’t post questions about a test or quiz that is found in school…

Explanation:

5 0
3 years ago
Read 2 more answers
PLEASE HELP WITH THESE!! I’m stuck
ludmilkaskok [199]
3. B
4. A
5.A
6. A (i think)
Hope this helps
4 0
3 years ago
5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------> 1 Mg3(
professor190 [17]

Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

y = (3×9.87)/2 = 14.80 moles

So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

Molar mass of Mg(OH)₂ = 58.3197 g/mol

Mass of Mg(OH)₂ required for the reaction

= 14.8 × 58.3197 = 863.13 g

Hope this Helps!!!

5 0
3 years ago
Read 2 more answers
Question 1<br> 1 pts<br> How many mols of bromine are present in 35.7g of<br> Tin(IV) bromate?
sleet_krkn [62]

Answer:

n = 0.0814 mol

Explanation:

Given mass, m = 35.7g

The molar mass of Tin(IV) bromate, M = 438.33 g/mol

We need to find the number of moles of bromine. We know that,

No. of moles = given mass/molar mass

So,

n=\dfrac{35.7}{438.33}\\\\n=0.0814\ mol

So, there are 0.0814 moles of bromine in 35.7g of  Tin(IV) bromate.

3 0
3 years ago
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