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tatyana61 [14]
3 years ago
12

What type of molecule is shown below

Chemistry
1 answer:
Kobotan [32]3 years ago
4 0

Answer is: C alkane.

Name of this molecule is n-hexane.

Hexane is alkane (acyclic saturated hydrocarbon, carbon-carbon bonds are single) of six carbon atoms.

Carbon atoms in hexane have sp3 hybridization.

In sp3 hybridization hybridize one s-orbital and three p-orbitals of carbon atom.

Alkene has one double bond and alkyne has one triple bond.

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Sugar can be converted to carbon dioxide and water by burning in air or by metabolic reactions inside living cells. How are thes
rjkz [21]

Answer:

These reactions are similar because the process is similar and the products are carbon dioxide they are different because the substances are different to outgo these reactions

Explanation:

5 0
3 years ago
How many grams of H 2O are produced from 28.8 g of O 2? (Molar Mass of H 2O = 18.02 g) (Molar Mass of O 2=32.00 g) 4 NH 3 (g) +
krek1111 [17]

Answer:  13.9 g of H_2O will be produced from the given mass of oxygen

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} O_2=\frac{28.8g}{32.00g/mol}=0.900moles

The balanced chemical reaction is:

4NIO_2(g)+7O_2(g)\rightarrow 4NO_2(g)+6H_2O(g)

According to stoichiometry :

7 moles of O_2 produce =  6 moles of H_2O

Thus 0.900 moles of O_2 will produce =\frac{6}{7}\times 0.900=0.771moles  of H_2O

Mass of H_2O=moles\times {\text {Molar mass}}=0.771moles\times 18.02g/mol=13.9g

Thus 13.9 g of H_2O will be produced from the given mass of oxygen

5 0
2 years ago
How many milliliters of 4.00 M NaOH are required to exactly neutralize 50.0 milliliters of a 2.00 M solution of HNO3 ?
kramer

Answer: The volume of NaOH required is 25.0 ml

Explanation:

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity HNO_3 = 1

M_1 = molarity of HNO_3 solution = 2.00 M

V_1 = volume of  HNO_3 solution = 50.0 ml

n_2 = acidity of NaOH = 1

M_1 = molarity of NaOH solution = 4.00 M

V_1 = volume of  NaOH solution =  ?

Putting in the values we get:

1\times 2.00\times 50.0=1\times 4.00\times V_2

V_2=25.0ml

Therefore, volume of NaOH required is 25.0 ml

3 0
2 years ago
Hat volume of a 0.540 M NaOH solution contains 12.5 g NaOH
amid [387]

Answer:

At 0.58 L of 0.540 M NaOH solution contain 12.5 g NaOH.

Explanation:

Given data:

At volume = ?

Mass of NaOH = 12.5 g

Molarity of solution = 0.540 M

Solution:

First of all we will calculate the number of moles of sodium hydroxide.

Number of moles = mass/molar mass

Number of moles = 12.5 g / 40 g/mol

Number of moles = 0.3125 mol

Volume of NaOH:

Molarity = number of moles / volume in L

Now we will put the values.

0.540 M = 0.3125 mol / volume in L

volume in L = 0.3125 mol / 0.540 mol/L

volume in L = 0.58 L

5 0
2 years ago
Chlorine has two stable isotopes, Cl-35 and Cl-37. If their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is
natima [27]

Answer:

X(Cl-35) = 75.95% => Answer 'A'

Explanation:

34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance

X(Cl-35) + X(Cl-37) = 1  ⇒  X(Cl-37) = 1 - X(Cl-25)

34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45

34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45

Rearrange ...

36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45

2.0006·X(Cl-35) = 1.5195

X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance

⇒ % abundance = 75.95%

3 0
3 years ago
Read 2 more answers
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