What kind of change is a change in state of matter and why Physical or chemical

Chemistry

1 answer:

Maru [420]3 years ago

6 0

Answer:

physical

Explanation:

Common changes of state include melting, freezing, sublimation, deposition, condensation, and vaporization. I hope this helps. :)

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In a chemical reaction, exactly 2 mol of substance A react to produce exactly 3 mol of substance B. 2 A ⟶ 3 B How many molecules

Harman [31]

Answer:

There will be produced 1.71 moles of B which contain 1.03×10²⁴ molecules

Explanation:

The example reaction is:

2A → 3B

2 moles of A produce 3 moles of B

If we have the mass of A, we convert it to moles and then, we make the rule of three: 29.2 g / 25.6g/mol = 1.14 moles

Therefore 2 moles of A produce 3 moles of B

1.14 moles of A will produce (1.14 . 3) / 2 = 1.71 moles of B are produced

Now we can determine, the number of molecules

1 mol has NA molecules (6.02×10²³)

1.71 moles have (1.71 . NA) = 1.03×10²⁴ molecules

4 0

3 years ago

What happens to the partial pressure of oxygen in a sample of air if the temperature is increased?

ludmilkaskok [199]

Air is mainly composed of N2 (78%), O2 (21%) and other trace gases. Now, the total pressure of air is the sum of the partial pressures of the constituent gases. The partial pressure of each gas, for example say O2, can be expressed as:

p(O2) = mole fraction of O2 * P(total, air) ----(1)

Thus, the partial pressure is directly proportional to the total pressure. If we consider a sealed container then, as the temperature of air increases so will its pressure. Based on equation (1) an increase in the pressure of air should also increase the partial pressure of oxygen.

6 0

3 years ago

Read 2 more answers

7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$