CHEMISTRY HIGH SCHOOL

At which temperature and pressure would a sample of helium behave most like an ideal gas?1
75 K and 500 kPa
2
150. Kand 500 kPa
3
300. Kand 50 kPa
4
600. K and 50 kPa
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Answers

Answer 1
Answer:

600k as well as 50kPa since the pressure seems to be the lowest as well as the temperature is the greatest throughout this situation.

Under high temperatures and low pressures, actual gases behaved like ideal gases. such as the rule

  • PV=nRT

Even though at high temperatures, the kinetic energy of such gas particles surpasses the intermolecular attractive forces, — in other words this same potential energy of something like the molecules (gaseous).

Thus the above answer i.e., "option 4" is right.

Learn more:

brainly.com/question/1371338
Answer 2
Answer:

Answer: 600K and 50kPa

Explanation:

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How will an energy diagram look for an exothermic reaction?The reactants and products will be at the same potential energy.



The reactants are at a higher potential energy compared to the products.



The reactants are at a lower potential energy compared to the products.



There will not be an activation energy barrier.

Answers

Answer : The correct option is, The reactants are at a higher potential energy compared to the products.

Explanation :

Exothermic reaction means energy is released and the overall enthalpy change is negative.

In exothermic reaction, the reactants are at a higher potential energy compared to the products. In other words, we can say that the products are more stable than the reactants.
B.) The reactants are at a higher potential energy compared to the products.

My reason being is because the reactants tend to gather "hyper" energy which is also known as potential energy and this will occur as shown in the answer as an Exothermic Reaction.

I really hope that this helps you out a lot. Have a nice day :)

Which equations shows the complete dissociation of a strong base? HCl -> cl- + H+

HNO2 -> H+ + NO2-

HA + h20 -> A- + H+ + h20

BOH + h20 -> B+ + OH- + H20

Answers

The correct answer is the fourth option. The complete dissociation of a strong base is BOH + h20 -> B+ + OH- + H20 since this is the only base from the choices given. A base is a substance that accepts hydrogen ions. 

Answer : The correct option is, (4) BOH+H_20\rightarrow B^++OH^-+H_20

Explanation :

Strong acid : Strong acids are those acids which are completely dissociates into ions in water. It dissociates into hydrogen ion and an anion.

Strong base : Strong base are those base which are completely dissociates into ions in water. It dissociates into hydroxide ion and a cation.

In all the given options, option (4) BOH+H_20\rightarrow B^++OH^-+H_20 equation shows the complete dissociation of a strong base.

While the other options shows the complete dissociation of an acid.

Although expensive, platinum is used as a catalyst to break down harmful gases in car exhaust into less harmful gases. Which statement best describes platinum in this use? a.It is a homogeneous catalyst, so only a small amount is needed even though it exits with the products.

b.It is a heterogeneous catalyst, so only a small amount is needed and it is easily separated from the products.

c.It is a homogeneous catalyst, so although a large amount is needed, it is easily separated from the products.

d.It is a heterogeneous catalyst, so a large amount is needed and it exits with the products.

Answers

Platinum is solid at room temperature. SInce the catalyst is solid while the medium involves gases, this catalyst is characterized as heterogenous. Because they are different states of matter, they can be easily separated through physical means. So, the answer is B.

Answer:

B EDGE 2022

Explanation:

at standard pressure how do the boiling point and the freezing point of NaCl compare to the boiling point and freezing point of H2O

Answers

Usually in this context you would be referring to the boiling and freezing point of a NaCl solution (saltwater) compared to pure H_{2}O. Sematics would be different for NaCl compound itself, you would say melting and boiling point for a solid substance- and the temperatures would be very, very radical (high). 
The boiling point of pure water is 100 degrees C (212 F), and the freezing/melting point is below 0 degrees C (32 F). For a salt water solution, the boiling point is raised and the melting point is lowered. This means that water will stay liquid for an increased range of temperature. Depending on the amount of NaCl solute in the water, the boiling and melting points may change a few degrees.

A 6.25mg sample of Cr51 decays for 111 days. After that amount of time, 0.75mg remains. What is the half-life of Cr51?

Answers

To find the half-life (\( t_{\text{half}} \)) of Cr51, we can use the radioactive decay formula:

\[ N = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{\text{half}}}} \]

Where:
- \( N \) is the remaining amount after time \( t \) (0.75 mg in this case)
- \( N_0 \) is the initial amount (6.25 mg)
- \( t \) is the time elapsed (111 days in this case)
- \( t_{\text{half}} \) is the half-life we're trying to find

We'll rearrange this formula to solve for \( t_{\text{half}} \):

\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{t_{\text{half}}}} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{N}{N_0}\right) = \frac{-t}{t_{\text{half}}} \times \ln\left(\frac{1}{2}\right) \]

Now, plug in the given values:

\[ \ln\left(\frac{0.75\, \text{mg}}{6.25\, \text{mg}}\right) = \frac{-111\, \text{days}}{t_{\text{half}}} \times \ln\left(\frac{1}{2}\right) \]

Solve for \( t_{\text{half}} \):

\[ t_{\text{half}} = \frac{-111\, \text{days}}{\ln\left(\frac{0.75\, \text{mg}}{6.25\, \text{mg}}\right) / \ln\left(\frac{1}{2}\right)} \]

Now, calculate this to find the half-life.

Using the provided formula to calculate the half-life (\( t_{\text{half}} \)), we get:

\[ t_{\text{half}} = \frac{-111\, \text{days}}{\ln\left(\frac{0.75\, \text{mg}}{6.25\, \text{mg}}\right) / \ln\left(\frac{1}{2}\right)} \]

Plugging in the values and calculating:

\[ t_{\text{half}} ≈ \frac{-111\, \text{days}}{-1.4978} \]
\[ t_{\text{half}} ≈ 74.24\, \text{days} \]

So, the half-life of Cr51 is approximately \( 74.24\, \text{days} \).

The vapor pressure of a liquid is 0.92 atm at 60°C. The normal boiling point of the liquid could be(1) 35°C (3) 55°C(2) 45°C (4) 65°C

Answers

The answer is (4) 65°C. For this question, you need to know that the definition of boiling is when a liquid's vapor pressure equals atmospheric pressure. Normal boiling point just means boiling point at sea level, when atmospheric pressure is 1atm. For this particular liquid, you would need to raise the temperature a little for the vapor pressure to rise to 1atm (because vapor pressure increases with temperature). Therefore your temperature must be higher than 60°C, and the only such choice available to you is (4).

Answer:

  • 4 ¥

TRUST REASON

The answer is 4 because when the vapor pressure is in liquid form, the boiling point would be the hottest point
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