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Komok [63]
3 years ago
7

Which substance represents a compound?(1) C(s) (3) CO(g)(2) Co(s) (4) O2(g)

Chemistry
2 answers:
zheka24 [161]3 years ago
7 0
Compound is the same thing as CO. 
Andrei [34K]3 years ago
5 0
C = Carbon
CO = Carbon Monoxide
Co = Cobalt
O2 = Oxygen

So the answer is (3) CO
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Identify the parts of the telescope.
Alona [7]
Eyepiece, finder-scope, optical tube, aperture, focuser, and mount
6 0
3 years ago
Read 2 more answers
mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of
anygoal [31]

<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

  • <u>For nitrogen gas:</u>

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

  • <u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0
3 years ago
The partial pressure of N2 in the air is 593 mm Hg at 1 atm. What is the partial pressure of N2 in a bubble of air a scuba diver
Evgen [1.6K]

Answer: Partial pressure of N_{2} at a depth of 132 ft below sea level is 2964 mm Hg.

Explanation:

It is known that 1 atm = 760 mm Hg.

Also,   P_{N_{2}} = x_{N_{2}}P

where,    P_{N_{2}} = partial pressure of N_{2}

                 P = atmospheric pressure

            x_{N_{2}} = mole fraction of N_{2}

Putting the given values into the above formula as follows.

      P_{N_{2}} = x_{N_{2}}P

    593 mm Hg = x_{N_{2}} \times 760 mm Hg

       x_{N_{2}} = 0.780

Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of N_{2} is as follows.

         P_{N_{2}} = x_{N_{2}}P

                  = 0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}

                  = 2964 mm Hg

Therefore, we can conclude that partial pressure of N_{2} at a depth of 132 ft below sea level is 2964 mm Hg.

8 0
3 years ago
How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?
DochEvi [55]

Answer:

1) 17.5 mL

Explanation:

Hello,

In this case, the reaction between sulfuric acid and potassium hydroxide is:

H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O

In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:

2*n_{acid}=n_{base}

And in terms of concentrations and volumes:

2*M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the volume of acid:

V_{acid}=\frac{M_{base}V_{base}}{2*M_{acid}} =\frac{35mL*1.5M}{2*1.5M} \\\\V_{acid}=17.5mL

Best regards.

5 0
3 years ago
What is the difference between the mass number for carbon- 14 and carbon's atomic number of 12.011 amu
Dennis_Churaev [7]
Great question, but I believe you are mixing up atomic number with mass number. Assuming you are, 12.011 amu is the average mass of a carbon atom. For carbon, it can come in three forms: carbon-12, carbon-13, carbon-14. The number following carbon is the mass number of that particular carbon "isotope". The reason the average is so close to 12 is because carbon-12 is by far the most common, so the average should be (and is) very close to 12. Therefore, 12.011 is a weighted average of all carbon molecules, and carbon-14 is a particular carbon molecule that weighs 14 amu. 
7 0
3 years ago
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