What effect does shape and size have on an object's density?

I need help on this please

tino4ka555 [31]

Answer:

The answer is c

Explanation:c

3 0

3 years ago

a sample of H2 is collected over water such that the combined hydrogen water vapor sample is held at a pressure of one standard

Trava [24]

Answer : The partial pressure of $H_2$ is 98.825 kPa.

Solution : Given,

Total pressure of hydrogen water vapor = 1 atm = 101.325 kPa (1 atm = 101.325 kPa)

Partial pressure of water vapor = 2.5 kPa

Dalton's law of partial pressure : It is defined as the total pressure exerted is equal to the sum of the partial pressure of individual gases.

Formula used :

$P_{Total}=\sum_{i=1}^nP_i$

$P_{Total}$ = Total pressure

$P_{i}$ = Partial pressure of individual gases

$P_{Total}=P_{H_2}+P_{\text{ Water vapor}}$

Now put all the given values in this expression, we get

$101.325kPa=P_{H_2}+2.5kPa$

$P_{H_2}=98.825kPa$

Therefore, the partial pressure of $H_2$ is 98.825 kPa.

7 0

3 years ago

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

viktelen [127]

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the $'n^{th}'$ orbit of hydrogen atom.

Formula used :

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV$

Energy of n = 2 in an hydrogen atom:

$E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV$

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

$E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV$

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

3 0

3 years ago

Which process is used by nuclear power plants to produce low cost, long lasting

lapo4ka [179]

Answer:

Nuclear fission is the one used

5 0

3 years ago

Which of the following are true statements about equilibrium systems?For the following reaction at equilibrium:2 H2(g) + O2(g) ?

VMariaS [17]

These are five questions about equilibrium systems each with its complete answer.

Question 1. For the following reaction at equilibrium:

2 H₂(g) + O₂(g) ⇄ 2 H2O(g), the equilibrium will shift to the left if the volume is doubled?

Answer: TRUE

Explanation:

When a force disturbs a chemical equillibrium, the system will shift toward the direction that reduces the effect. This is Le Chatelier's principle.

As per Bolye's law, at constant temperature, the volume and the pressure of a fixed amount of gas are inversely related.

Also, the pressure of the system is directly related to the number of particles (atoms or molecules). Hence, more molecules, more pressure; less molecules, less pressure.

Now, you can reason in this way: if the volume of the given system is doubled, then the pressure is lowered, and the system will try to alleviate this disturbance by shifting the reaction to the side that produces more molecules, to restore the pressure. Because on the left side three molecules can be produced from the reaction of two molecules of H₂O on the rihgt, the system will shift to the left. And this proves the truth of the statement.

Question 2. For the following reaction at equilibrium:

H₂(g) + F₂(g) ⇄ 2HF(g), removing H₂ will decrease the amount of F₂ present once equilibrium is reestablished.

Answer: FALSE.

Explanation:

Note that, since the temperature and other conditions have not changed, the equilibrium constant, Ke, has not changed. And, for the given equilibrium, Ke is given by the following equation.

Ke = [ H₂] [F₂] / [HF]²

Hence, to keep Ke unchanged, when removing H₂, the amount of F₂ present once equilibrium is reestablished will have to increase.

This is the opposite of the stated on the question, so the statement is false.

Question 3. Increasing the temperature of an exothermic reaction shifts the equilibrium position to the right.

Answer: FALSE.

Explanation:

You can write an exothermic equlibrium placing heat as a product on the right side of the equation; in this way:

A + B ⇄ C + D + heat

There, treating the heat as another product, you can reason that increasing the temperature, which is equivalent to supplying heat, will shift the equilibrium to the left side to consume heat, instead to the proposed by the statement. So, this is a false statement.

Question 4. For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), adding more CaCO₃ will shift the equilibrium to the right.

Answer: TRUE.

Explanation:

CaCO₃(g) is the only reactant of the forward reaction.

Adding more CaCO₃ may be seen as a disturbance against which the system will act by consuming it and producing more CaO and CO₂.

So, the forward reation will be favored and you conclude that adding more CaCO₃ will shift the equilibrium to the right.

Question 5. For the following reaction at equilibrium:

CaCO₃(s) ⇄ CaO(s) + CO₂ (g), increasing the total pressure by adding Ar(g) will have no effect on the equilibrium position.

Answer: TRUE.

Explanation:

In accordance to Le Chatelier's principle, increasing the pressure should be addresed by the equilibrium by shifting to the side where such pressure increase could be released.

That is possible when the number of molecules of gases on both sides are different: the equilibrium will shift to the side where more molecules less molecules are produced.

But, when the stoichiometry of the reaction shows the same number of molecules on both sides, which is the case in the given equilibrium, increasiing (or decreasing) the pressure will have no effect on the equilibrium position. Then, the answer is true.

8 0

3 years ago