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Llana [10]
3 years ago
11

A sample of table sugar (sucrose c12h22o11) has a mass of 3.115 g

Chemistry
1 answer:
attashe74 [19]3 years ago
6 0
Hey there:

a) atomic mass:

Carbon =<span>12.0107 g/mol

</span>Hydrogen = <span>1.00794 g/mol

Oxygen =  </span><span>15.9994 g/mol
</span>
Therefore:


C12H22O11 = 


12 * 12.0107 + 1 * 1.00794 + 16 * 15.9994 => <span>342.29648 g/mol

__________________________________________________________


b) number of moles:

n = m / mm

n = 3.115 / </span><span>342.29648 

n = 0.0091 moles

________________________________________________

hope this helps!</span>
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ludmilkaskok [199]
3. B
4. A
5.A
6. A (i think)
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4 0
3 years ago
How many atoms are in one molecule of baking soda, NaHCO3?
vladimir2022 [97]

Answer: 6 atoms in total

Explanation:

It has one sodium atom, one hydrogen atom, one carbon atom, and three oxygen atoms.

6 0
3 years ago
What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0
3 years ago
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i
omeli [17]

Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500 L

The concentration sodium sulfate = 2.104 M

Step 2: The equation

Na2SO4 → 2Na+ + SO4^2-

For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)

Step 3: Calculate the concentration of the ions

[Na+] = 2*2.104 M = 4.208 M

[SO4^2-] = 1*2.104 M = 2.104 M

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

8 0
3 years ago
The lowest energy state of an atom is its
erica [24]

Answer:

Principle quantum number.

Explanation:

8 0
3 years ago
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