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maria [59]
2 years ago
9

Chemistry balance equations C2H +O2=CO2+H2O​

Chemistry
1 answer:
son4ous [18]2 years ago
7 0

Invalid equation.

..............

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How do cells use lipids and carbohydrates?
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Answer:

Animals tend to use carbohydrates primarily for short-term energy storage, while lipids are used more for long-term energy storage. Carbohydrates are stored as glycogen in animals while lipids are stored as fats (in plants carbohydrates are stored as cellulose and lipids as oils)

Explanation:

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Which mixture can be classified as a homogeneous mixture?
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Anything can be homogenous as long as you can only see the same type of liquid
think about it like this
orange juice with pulp is Hetero
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Why is vsepr theory important?
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It is very important<span> to know the shape of a molecule if one is to understand its reactions. It is also desirable to have a simple method to predict the geometries of compounds. For main group compounds, the </span>VSEPR<span> method is such a predictive tool and unsurpassed as a handy predictive method.</span>
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Consider a 400 mL solution of 0.10 M \ce{NaOH}NaOH. Calculate the mass of solid \ce{NaOH}NaOH required to achieve this solution.
AnnyKZ [126]

Answer:

Is 40.10 ml

Explanation:

i've dome it before

3 0
2 years ago
An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu
chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

The density of a proton is 6.278\times 10^{14} g/cm^3.

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\pi r^3..[1]

Diameter of the nucleus ,d' = 9.00\times 10^{-5}\AA

Radius of the nucleus ,r' = 0.5 d'=0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA

Volume of nucleus = V'

V=\frac{4}{3}\pi r'^3..[2]

Dividing [2] by [1]

\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}

=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}

\frac{V'}{V}=4.6656\times 10^{-14}

The fraction of the volume of the atom that is taken up by the nucleus is 4.6656\times 10^{-14}.

Diameter of the proton ,d = 1.72\times 10^{-15} m = 1.72\times 10^{-13} cm

1 m = 100 cm

Radius of the proton,r = 0.5 d=0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm

Volume of the sphere= \frac{4}{3}\pi r^3

Volume of atom = V

V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3

Mass of proton, m = 1.0073 amu = 1.0073\times 1.66054\times 10^{-24} g

1 amu = 1.66054\times 10^{-24} g

Density of the proton : d

d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3

The density of a proton is 6.278\times 10^{14} g/cm^3.

5 0
3 years ago
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