Ask question

Ask question

All categories

3 years ago

5

Which of these are examples of physical changes? Check all that apply. burning wood freezing water cutting paper iron rusting mi

xing water and acetone

2 answers:

rodikova [14]3 years ago

5 0

Cutting paper
freezing water
mixing water
acetone

Colt1911 [192]3 years ago

4 0

<u>Answer:</u>

freezing water

cutting paper

mixing water and acetone

You might be interested in

1.6g of a compound were found to contain 0.64g of copper, 0.32g of sulfur and 0.64g of oxygen. Calculate it's empirical formula.

Tresset [83]

The empirical formula of the compound containing 0.64 g of copper, 0.32 g of sulphur and 0.64 g of oxygen is CuSO₄

From the question given above, the following data were obtained:

Mass of Cu = 0.64 g

Mass S = 0.32 g

Mass of O = 0.64 g

<h3>Empirical formula =? </h3>

The empirical formula of the compound can be obtained as follow:

Cu = 0.64 g

S= 0.32 g

O = 0.64 g

Divide by their molar mass

Cu = 0.64 / 63.5 = 0.01

S= 0.32 / 32 = 0.01

O = 0.64 / 16 = 0.04

Divide by the smallest

Cu = 0.01 / 0.01 = 1

S= 0.01 / 0.01 = 1

O = 0.04 / 0.01 = 4

Therefore, the empirical formula of the compound is CuSO₄

Learn more: brainly.com/question/13086832

6 0

3 years ago

The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at

horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0

4 years ago

In the reaction 2NO (g) + O2 (g) --> 2NO2 (g) . Which species has the highest rate of consumption (decrease)?

Genrish500 [490]

Answer:

NO.

Explanation:

Because it is consumed by a factor of 2 moles as the reaction proceeds.

5 0

3 years ago

Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). Match Term Definition Incr

KIM [24]

Answer:

1) Increasing the pressure C) Shift to the right

2) Removing hydrogen gas A) Shift to the left

3) Adding a catalyst B) No effect

Explanation:

<em>Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>

<em />

<u><em>1) Increasing the pressure:</em></u>

When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.

The reactants side (left) has 4.0 moles of gases and the products side (right) has 2.0 moles of gases.

So, increasing the pressure will shift the reaction to the side with lower moles of gas (right side).

<u><em>so, the right match is: C) Shift to the right.</em></u>

<em><u>2) Removing hydrogen gas:</u></em>

Removing hydrogen gas will decrease the concentration of the products side, so the reaction will be shifted to the lift side to suppress the decrease in the concentration of hydrogen gas by removing.

<em>so, the right match is: A) Shift to the left.</em>

<u><em>3) Adding a catalyst:</em></u>

Catalyst increases the rate of the reaction without affecting the equilibrium position.

Catalyst increases the rate via lowering the activation energy of the reaction.

This can occur via passing the reaction in alternative pathway (changing the mechanism).

The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).

in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.

with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.

<em>so, the right match is: B) No effect.</em>

4 0

3 years ago

Using the periodic table and your knowledge of atomic structure. Compare the number of electrons in a carbon-12 and carbon-14

Brut [27]

Both carbon 12 and carbon 14 have the same no. of Electrons but the difference is carbon 14 has more neutrons

6 0

3 years ago

Read 2 more answers