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4 years ago

Which of these can be classified as a substance A) oxygen B) carbon dioxide C) air D) acid rain

Chemistry

2 answers:

ratelena [41]4 years ago

4 0

D could be a answer

Julli [10]4 years ago

4 0

Answer:

carbon dioxide

Explanation:

the rest are mixtures

A chemical substance is composed of one type of atom or molecule.

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Annabelle was explaining the carbon cycle to her friend. She said that all the carbon

vagabundo [1.1K]

Answer: O2+6H12O6=CO2+ENERGY(ATP)

I DON'T THINK SHE IS CORRECT

Explanation:

5 0

3 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

3 years ago

275 g of a compound with the molecular formula ccl2f2 contains how many molecules of the compound

madreJ [45]

brainly.com/question/8581600#respond

3 0

3 years ago

What is the mass percent of oxygen (0) in SO2?

vazorg [7]

Answer:

D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

Explanation:

Step 1: Detemine the mass of O in SO₂

There are 2 atoms of O in 1 molecule of SO₂. Then,

m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g

Step 2: Determine the mass of SO₂

m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g

Step 3: Detemine the mass percent of oxygen in SO₂

We will use the following expression.

m(O)/m(SO₂) × 100%

(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

5 0

3 years ago

A gas fills a balloon at a temperature of 27 C and 101.325 kPa of pressure. What will the pressure of the balloon be if the gas

JulijaS [17]

135.1‬kPa

Explanation:

Given parameters:

T1 = 27°C

P1 = 101.325 kPa

T2 = 127°C

Unknown:

P2 = ?

Solution:

Using a derivative of the combined gas law where we assume that the gas has a constant volume, we can solve for the unknown.

At constant volume:

$\frac{P1}{T1} = \frac{P2}{T2}$

P1 is the initial pressure

T1 is the initial temperature

P2 is the final pressure

T2 is the final temperature

Take the given temperature to K

T1 = 27 + 273 = 300K

T2 = 127 + 273 = 400K

Input the variables:

$\frac{101.325}{300} = \frac{P2}{400}$

P2 = 135.1‬kPa

learn more:

Boyle's law brainly.com/question/8928288

#learnwithBrainly

3 0

3 years ago