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3 years ago

Which of the following does not represent a compound? NaCl N20 CO2 O2

Chemistry

1 answer:

gavmur [86]3 years ago

7 0

Answer:

Explanation:

A compound is made of two or more chemical elements. With that in mind, O2 is only one element: oxygen.

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Element A has 15 protons. Determine the number of neutrons in its isotope with mass number 33.

bagirrra123 [75]

In an electrically neuteral atom, number of protons = number of electrons = atomic number.
Mass number = neutrons + protons/electrons/atomic number
Therefore,
neutrons = mass number - <span>protons/electrons/atomic number
Neutrons = 33 - 15 = 18

The answer is thus B. But this is the solution and explanation along with it as proof.</span>

6 0

3 years ago

Why does a microscope stage have a small hole in it?

jarptica [38.1K]

Answer: B. Allow light to pass through. :)

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3 years ago

The sulfur atom of sulfur dioxide is considered to be sp2-hybridized. The expected bond angle is 120°, but is actually slightly

Alja [10]

Answer: The bonds are intermediate between double and single bonds

Explanation:

A closer look at the diagram below shows that the bonds in sulphur IV oxide are intermediate between double and single bonds. Hence they do not have the exact bond angle of single bonds. This is why the bond angle is not exactly 120°. There are two resonance structures in the diagram that clearly show this point.

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3 years ago

Which conditions must be met for two triangles to be similar

lbvjy [14]

The included angle, i.e. If two pairs of sides of two triangles are in proportion, and the included angles are equal, then the triangles are similar.

7 0

3 years ago

The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k

Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

$ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

We have given in the question

$P_1=102\ mmHg$

$T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol$

And $R$ is the Universal Gas Constant.

$R=0.008 314 kJ/Kmol$

$ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165$

Taking inverse log both side we get,

$\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg$

8 0

3 years ago