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3 years ago

How many hydrogen atoms are in 1 molecule of C3H7O

Chemistry

2 answers:

Zarrin [17]3 years ago

8 0

Answer:

Explanation:

it has no tiny number on it

Ostrovityanka [42]3 years ago

3 0

Answer:7

Explanation:5

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The following information was recorded by a student team working to prepare nickel sulfate. Plan: Prepare NiSO4 by reacting NiO

mixas84 [53]

Answer:

The options e and d are correct.

Explanation:

Mass of NiO = 7.5 g

Moles of NiO = $\frac{7.5 g}{74.69 g/mol}=0.10 mol$

Moles of sulfuric acid = n

Volume of sulfuric acid ,V= 50 mL = 0.050 L

Molarity of sulfuric acid ,M = 6 mol/L

$n=M\time V=6mol/L\times 0.050 L =0.3 mol$

$NiO + H_2SO_4\rightarrow NiSO_4 + H_2O$

According to reaction, 1 mole of NiO reacts with 1 mole of sulfuric acid.

Then 0.10 moles of NiO reacts with :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of sulfuric acid.

As we can see that sulfuric acid is in excess amount, so the amount of the product will depend upon amount of NiO.

According to reaction, 1 mole of NiO gives with 1 mole of $NiSO_4$ .

Then 0.10 moles of NiO wil give :

$\frac{1}{1}\times 0.10 mol/=0.10 mol$ of $NiSO_4$ .

Molar mass of $NiSO_4$ = 154.75 g/mol

Mass of 0.10 moles of $NiSO_4$ :

= 154.75 g/mol × 0.10 mol = 15.475 g

Theoretical mass of $NiSO_4$ = 15.475 g

Experimental yield of $NiSO_4$ = 17.4 g

Percentage yield :

$\Yield=\frac{\text{Experimental mass}}{\text{Theoretical mass}}\times 100$

Percentage yield of $NiSO_4$ :

$\Yield=\frac{17.4}{15.475 g}\times 100=112\%$

Moles of $NiSO_4.6H_2O$ = 262.85 g/mol × 0.10 mol = 26.285 g

Experimental yield of $NiSO_4.6H_2O$ = 17.4 g

Percentage yield of $NiSO_4.6H_2O$ :

$\Yield=\frac{17.4}{26.285 g}\times 100=66.2\%$

3 0

3 years ago

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1o) = carbon bonded to just 1 other ca

Alla [95]

The question is incomplete, complete question is :

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?

Structure is given in an image?

Answer:

There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

Explanation:

Total numbers of carbon = 10

Number of primary carbons that is carbon joined to just single carbon atom = 6

Number of secondary carbons that is carbon joined to two carbon atoms = 1

Number of tertiary carbons that is carbon joined to three carbon atoms = 2

Number of quartenary carbons that is carbon joined to four carbon atoms = 1

So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

3 0

3 years ago

What type of bond would form between two atoms of selenium?

siniylev [52]

Answer:

the correct answer is B

hope it helps !!!

8 0

3 years ago

Identify the types of chemical reactions occurring in the chemical equations.

Inga [223]

Answer: science

Explanation:

3 0

3 years ago

Determine the numbers of atoms in each of the following 5.40 g B 0.250 mol k 0.0384 mol k 0.02550 g pt 1.00 x 10^-10 g Au,

Pani-rosa [81]

The answers are the following:
1. 3.01 x 10^23 atoms B
solution: (5.40/10.81)(6.022x10^23)
2. 1.51 x 10^23 atoms S
solution: (.250)(6.022x10^23) 
3. 2.31 x 10^22 atoms K
solution: (.0384)(6.022x10^23) 
4. 7.872 x 10^19 atoms Pt
solution: (.02550/195.08)(6.022x10^23) 
5. 3.06 x 10^11 atoms Au
solution: (1.00x10^-10/196.97)(6.022x10^23)

8 0

3 years ago

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