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3 years ago

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]

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

Explanation:

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:

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

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

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

For hydrogen gas:

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

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

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

For hydrogen gas:

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