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3 years ago

The arrows in the chart below represent phase transitions.

Chemistry

1 answer:

Lana71 [14]3 years ago

4 0

Answer:

a. 1,2,and 3.

Explanation:

When something turns liquid, e.g. metal, you need to heat it. When something turns gas, e.g. water, you need to heat it. And when something goes directly from solid to gas, e.g. dry ice, carbon dioxide, you need to add heat to it. hope this helps! Please mark brainiest.

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A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga

vivado [14]

Answer:

The molar mass of the unknown gas is 100.4 g/mol

Explanation:

Step 1: Data given

Molar mass of argon = 39.95 g/mol

After filling with argon the flask gained 3.221 grams

After filling with an unknown gas, the flask gained 8.107 grams

Step 2: Calculate the molar mass of the unknown gas

The gas with the higher molar mass will have the higher density.

Ar - 3.224 g; molar mass = 39.95 g/mol

X = 8.102 g; molar mass = ??

Molar mass of the unknown gas = 8.102g X *(39.95 g/mol / 3.224 g) = 100.4 g/mol

The molar mass of the unknown gas is 100.4 g/mol

8 0

3 years ago

Pls help fast!!! Test due soon!

Firlakuza [10]

Answer:

Explanation:

your wellcome

4 0

3 years ago

Read 2 more answers

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

4 years ago

statement 1,mass is conserved durning the reaction. statement 2 sum of mass and energy is conserved durning the reaction. which

maksim [4K]

Answer:

Explanation:

None of the statement is true for both chemical and nuclear reactions. In chemical reactions, mass is always conserved and the type of atoms are also conserved.

6 0

3 years ago

126785033% Divided by 348675948849 = <br> what does it equal

meriva

Answer:

0.00000363618

could be wrong.

double check me someone or just trust me

(don't blame me if you get it wrong)

5 0

3 years ago