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professor190 [17]
2 years ago
14

Calculate the empirical formula of a compound that is 56.27% phosphorus and the remainder is oxygen.

Chemistry
1 answer:
Inessa [10]2 years ago
5 0

Answer:

ikjnfjngvfnjfdkmldskmofevivrtbhurt3gdkjbfongidfngherghihierihgdifngodfnvgndfugiripegndbjgindgndfvcl vncf 'g

Explanation:

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Identify the reactant and products in the following chemical reaction: CH4 + 2O2 ➡ CO2 + 2H2O
kap26 [50]

The reactants are methane and oxygen.

The products are carbon dioxide and water.

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3 years ago
What was George Washington Caver professional information ​
Sedbober [7]

Explanation:

George Washington Carver was born enslaved and went on to become one of the most prominent scientists and inventors of his time, as well as a teacher at the Tuskegee Institute. Carver devised over 100 products using one major crop — the peanut — including dyes, plastics and gasoline.

5 0
3 years ago
Read 2 more answers
When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?
Ad libitum [116K]

You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.

For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.

In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!

6 0
3 years ago
Please help 16-20 thanks
MaRussiya [10]
16.  FALSE
17. TRUE
18. FALSE
19. TRUE
20.  TRUE
3 0
3 years ago
You are given 3.0 grams of solid sodium to react to pure water which has a molarity of 55.6 M. How many milligrams of H2 can be
serious [3.7K]

Answer:

= 15.51 mL

Explanation:

Here's is the reaction:

2HgO(s) ⇒ 2 Hg(s)+O₂(g)

In this reaction 2mol HgO =  1mol O₂

The molecular weight of HgO = 216.59g

so, 3.0g HgO = 3.0g x 1.00molHgO/216.59gHgO

= 0.0138511 molHgO

The amount of Oxygen follows:

0.0138511 molHgOx1/2= 0.00692555 mol O₂

Now, volume of 1 any gas = 22400mL

so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂

= 15.513232mL O₂

4 0
3 years ago
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