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laila [671]
3 years ago
7

PLEASE HELP ITS A TESTT!!

Chemistry
1 answer:
Sergio [31]3 years ago
6 0

Answer:

D

Explanation:

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Air is a solution and a mixture of gases, what is the solvent in the solution we call air?
MariettaO [177]

Answer:

c. nitrogen gas

Explanation:

Our air is a homogeneous mixture of many different gases and therefore qualifies as a solution. Approximately 78% of the atmosphere is nitrogen, making it the solvent for this solution.

8 0
3 years ago
Which of the following is an example of a chemical change? A. Boiling water B. Burning paper C. Dissolving sugar in water D. Mel
SpyIntel [72]
It’s B.burning paper
8 0
4 years ago
Read 2 more answers
Please help me :,) on 5 and 6
alexandr402 [8]

Answer: ??????????????????? huh

Explanation:

6 0
3 years ago
50 POINTS!
Varvara68 [4.7K]
5.5/-38 / 5.5 = 1/-6.9 x 1.3 = 1.3/-8.98
unit volume/temperature x searching amount

i’d say the temperature would be -8.98 C simply - I don’t know what formula youd use for this

6 0
4 years ago
Calculate by (a)% weight and (b) %mole each of the elements present in sugar
Musya8 [376]

Explanation:

Molecular mass of sugar = C_{12}H_{22}O_{11} : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100

Percentage of carbon by weight in C_{12}H_{22}O_{11}:

\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%

Percentage of hydrogen by weight in C_{12}H_{22}O_{11}:

\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%

Percentage of oxygen by weight in C_{12}H_{22}O_{11}:

\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%

b) Percentage of mole each of the elements present in sugar:

=\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in C_{12}H_{22}O_{11}:

\frac{12 mol}{45 mol}\times 100=26.66\%

Percentage of hydrogen by mole in C_{12}H_{22}O_{11}:

\frac{22 mol}{45 mol}\times 100=48.88\%

Percentage of oxygen by mole in C_{12}H_{22}O_{11}:

\frac{11 mol}{45 mol}\times 100=24.44\%

7 0
4 years ago
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