At 27.0°C, the volume of a gas is 630 L. At the same pressure, its volume is 92,0 mL at a temperature of

Answers

Answer 1

Answer:

–272.96 °C

Explanation:

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

Initial temperature (T₁) = 27.0 °C

Initial volume (V₁) = 630 L.

Final volume (V₂) = 92.0 mL

Final temperature (T₂) =?

Next, we shall convert 27.0 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 27.0 °C

Initial temperature (T₁) = 27.0 °C + 273

Initial temperature (T₁) = 300 K

Next, we shall convert 92.0 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

92 mL = 92 mL × 1 L / 1000 mL

92 mL = 0.092 L

Next, we shall determine the final temperature.

Initial temperature (T₁) = 300 K

Initial volume (V₁) = 630 L.

Final volume (V₂) = 0.092 L

Final temperature (T₂) =?

V₁ / T₁ = V₂ / T₂

630 / 300 = 0.092 / T₂

2.1 = 0.092 / T₂

Cross multiply

2.1 × T₂ = 0.092

Divide both side by 2.1

T₂ = 0.092 / 2.1

T₂ = 0.04 K

Finally, we shall convert 0.04 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

Final temperature (T₂) = 0.04 K

Final temperature (T₂) = 0.04 – 273

Final temperature (T₂) = –272.96 °C


Related Questions

explain why hydrogen chloride does not conduct electricity, but a solution of hydrogen chloride and water conduct electricity

Answers

The answer is… Ions are highly charged species that can readily conduct an electrical current. But hydrogen chloride the gas does not exist as ions since there is no solvent medium for them to dissolve into. Hope this helps!

Inquiry Extension Consider a reaction that occurs between solid potassium and chlorine gas. If you start with an initial mass of 15.20 g K, and an initial mass of 2.830 g Cl2, calculate which reactant is limiting. Explain how to determine how much more of the limiting reactant would be needed to completely consume the excess reactant. Verify your explanation with an example

Answers

The 3.13 g of K would be needed to completely react with the remaining [tex]Cl_2[/tex].

To determine which reactant is limiting, we need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces less product is the limiting reactant, since the reaction cannot proceed further once it is consumed.

The balanced chemical equation for the reaction between solid potassium and chlorine gas is:

2 K(s) + [tex]Cl_2[/tex](g) -> 2 KCl(s)

From the equation, we can see that 2 moles of K react with 1 mole of [tex]Cl_2[/tex] to form 2 moles of KCl.

First, we need to convert the masses of K and [tex]Cl_2[/tex] into moles:

moles of K = 15.20 g / 39.10 g/mol = 0.388 mol

moles of [tex]Cl_2[/tex] = 2.830 g / 70.90 g/mol = 0.040 mol

Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of KCl from each reactant:

Theoretical yield of KCl from K: 0.388 mol K x (2 mol KCl / 2 mol K) = 0.388 mol KCl

Theoretical yield of KCl from [tex]Cl_2[/tex]: 0.040 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) = 0.080 mol KCl

We can see that the theoretical yield of KCl from K is 0.388 mol, while the theoretical yield of KCl from [tex]Cl_2[/tex] is 0.080 mol. Therefore, the limiting reactant is [tex]Cl_2[/tex], since it produces less product.

To determine how much more of the limiting reactant would be needed to completely consume the excess reactant, we can use the stoichiometry of the balanced equation.

We know that 1 mole of [tex]Cl_2[/tex] reacts with 2 moles of K to produce 2 moles of KCl. Therefore, the amount of additional K needed to react with the remaining [tex]Cl_2[/tex] can be calculated as follows:

moles of K needed = 0.040 mol [tex]Cl_2[/tex] x (2 mol K / 1 mol [tex]Cl_2[/tex])

                                = 0.080 mol K

This means that 0.080 moles of K would be needed to completely consume the remaining [tex]Cl_2[/tex]. We can convert this to a mass by multiplying by the molar mass of K:

mass of K needed = 0.080 mol K x 39.10 g/mol

                              = 3.13 g K

Therefore, The 3.13 g of K would be needed to completely react with the remaining.

Example verification:

Suppose we had an additional 0.50 g of [tex]Cl_2[/tex] in the reaction. Would all of the K be consumed, or would there still be excess K?

Moles of additional [tex]Cl_2[/tex] = mass of [tex]Cl_2[/tex] / molar mass of [tex]Cl_2[/tex]

Moles of additional [tex]Cl_2[/tex] = 0.50 g / 70.90 g/mol

Moles of additional [tex]Cl_2[/tex] = 0.0070 mol

The theoretical yield of KCl that can be formed from the additional [tex]Cl_2[/tex] is:

0.0070 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) x (74.55 g KCl / 1 mol KCl) = 1.04 g KCl

Therefore, the total amount of KCl that can be formed from all of the [tex]Cl_2[/tex] is:

5.95 g + 1.04 g = 6.99 g

The amount of K that would be needed to completely consume all of the [tex]Cl_2[/tex].

Learn more about Solid Potassium at

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Cell membranes are selectively permeable. This means that A. only water can move freely across the cell membrane. B. any substance can move across the cell membrane, but chemical energy will always be required. C. some substances can move freely across the cell membrane, while others must be transported. D. no substances can move freely across the cell membrane.

Answers

Answer:

C. some substances can move freely across the cell membrane, while others must be transported.

Explanation:

A balloon is inflated to a volume of 8.0 L on a day when the atmospheric pressure is 1.013 bar . The next day, a storm front arrives, and the atmospheric pressure drops to 0.968 bar . Assuming the temperature remains constant, what is the new volume of the balloon, in liters

Answers

Answer:

[tex]V_2=8.4L[/tex]

Explanation:

Hello there!

In this case, according to the definition of the Boyle's law, which describes de pressure-volume behavior as an inversely proportional relationship, it is possible for us to write:

[tex]P_1V_1=P_2V_2[/tex]

Thus, since we are given the initial pressure and temperature, and the final pressure, we are able to calculate the final volume as shown below:

[tex]baV_2=\frac{P_1V_1}{P_2}\\\\V_2=\frac{8.0L*1.013bar}{ 0.968bar}\\\\V_2=8.4L[/tex]

Regards!

when rolling a number cube 500 times, how many times you expect to get a 3?

Answers

Answer:

[tex]\frac{250}{3}[/tex]

Explanation:

you can expect to get a 3 (theoretically) 1 time every 6 times you roll. A 1/6 chance.

Here's the equation:

[tex]\frac{1}{6} =\frac{x}{500}[/tex]

cross multiply (i think that's what it is called)

500=6x

divide by 6 on both sides:

x=[tex]\frac{250}{3}[/tex] or approx 83 times.

Hope this helps! Lmk if u have more questions <3

In calorimetry, energy is measured through heat transfer from one substance to
another. Which of the following is NOT a method of heat transfer?

Answers

I think you forgot to add a picture?

Answer:

Refraction

Explanation:

There are four stages to the classical demographic transition model Pre-transitional Europe was characterized by high and
fluctuating mortality and a high birth rate. The transition model began to progress into and through stage 2 in the late 18th and early
19th century. All BUT ONE contributed to the decline in mortality.
S- -1]))
A)
Enacting measures to provide clean water supplies.
B)
Public health advances including quarantine of settlements undergoing
epidemics
The development of vaccines to prevent disease and antibiotics to treat
infection.
D)
Widespread acceptance of germ theory resulting in more hygienic
practices, including hand washing and sterilizing medical equipment and
infants' bottles.

Answers

The answer is D! Explaination:

write half-reactions that show how H2O2 can act as either an oxidizing agent or a reducing agent, and describe where each of these situations occurred in your testing.

Answers

Answer:

H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .

Explanation:

When

H2O2 acts as an oxidizing agent

H2O2 + 2e- 2H+--->   2H2O

Reducing agent

H2O2 --> O2 + 2e + 2H+

H2O2 reduces itself to H2O and also oxidizes to O2 simultaneously thereby acting both as an oxidizing and reducing agent .

What does the cell theory state? Answer F All organisms are composed of a nucleus G All prokaryotes are composed of multiple cells H All prokaryotes are single celled organisms J All organisms are composed of cells

Answers

Answer:

(J) All organisms are composed of cells


A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0°C to 28.5 °C. Find the
mass of the water.

Answers

Answer:

1.5 × 10³ g

Explanation:

Step 1: Given and required data

Transferred heat (Q): 41,840 JInitial temperature: 22.0 °CFinal temperature: 28.5 °CSpecific heat capacity of water (c): 4.184 J/g.°C

Step 2: Calculate the temperature change

ΔT = 28.5°C - 22.0 °C = 6.5 °C

Step 3: Calculate the mass (m) of water

We will use the following expression.

Q = c × m × ΔT

m = Q / c × ΔT

m = 41,840 J / (4.184 J/g.°C) × 6.5 °C = 1.5 × 10³ g

Calculate how many grams of methane (CH4) are in a sealed 800. mL flask at room temperature (22 °C) and 780. mm of pressure. Show work pls.

Answers

"0.0340" mol of CH₄ are in sealed flask.

Methane (CH₄)

Methane would also be a greenhouse gas, therefore its existence tends to affect humanity's surface temp as well as weather patterns framework; it is released into the atmosphere from such a wide assortment of life forms as well as biogenic.

According to the question,

Volume, V = 800 mL or, 0.800 L

Temperature, T = 22°C or, 295

Pressure, P = [tex]\frac{780}{760}[/tex] = 1.03 atm

As we know the relation,

The gram of moles will be will be:

→ n = [tex]\frac{PV}{RT}[/tex]

By substituting the values, we get

     = [tex]\frac{1.03\times 0.800}{0.08206\times 295}[/tex]

     = [tex]\frac{0.824}{242.077}[/tex]

     = 0.0340

Thus the response above is correct.

Find out more information about methane here:

https://brainly.com/question/24690118

Calculate the volume of solvent present in a 55.5%
by volume of 10.5 mL alcohol solution.

Answers

Answer:

I dont know

Explanation:

good luck

The equivalence point of a titration corresponds to which of the following?
O the point where equal volumes of acid and base have been used
O Equivalence point is another term for end point
All of the listed options are true
Equivalence point is defined as the point where the pH indicator changes color
O the point where the acid and base have been added in proper stoichiometric amounts

Answers

Answer:

E: the point where the acid and base have been added in proper stoichiometric amounts

Explanation:

Equivalence point in titration is simply the point where the amounts of acid and base used just sufficiently reacts chemically to cause neutralization whereas the endpoint is the point where the indicator of the titration changes colour.

The Equivalence point occurs before the endpoint.

Thus, option E is correct.

PLEASE HELP I HAVE 19 MINUTES LEFT I WILL MARK BRAINLIEST
How much more acidic is a pH of 4 as compared to a pH of 6.5?

Answers

Answer:

316.227766

Explanation:

Answer 3.16 hope it helps

1. How does a virus differ from a common cell?
A. It has no nucleus, cell wall, or organelles.
B. It has two nuclei and no cell wall or organelles.
C. A virus has no cell well, no nucleus, and only organelles for
movement.
D. A virus differs from a cell only in shape.

Answers

The answer is letter C

A student weighs 0.347 g of KHP on a laboratory balance. The KHP was titrated with NaOH and the concentration of the NaOH determined to be 0.110 M. For the second titration, the student correctly diluted 6 M HCl from the reagent shelf using a graduated cylinder to obtain approximately 0.6 M HCl. This solution was titrated with the original NaOH solution. The student calculated the concentration of NaOH from the experiment to be 0.099 M. In which experiment should the student be more confident of the concentration of the NaOH solution

Answers

Answer:

Following are the solution to the given question:

Explanation:

Each method through KHP is somewhat more precise since we have weighed that requisite quantity, we exactly know the KHP intensity appropriately. Its initial 6 M HCl concentration was never considered mandatory. They have probably prepared 6 M HCl solution although long ago and could have changed its concentration over even a period.

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