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olasank [31]
2 years ago
12

A gas has a pressure of 3 atm at 350°C. What will its pressure be at 250°C? The volume and amount of gas is constant. Hint: Conv

ert temperature to Kelvin before calculating (°C + 273 = K)
\
Chemistry
1 answer:
vichka [17]2 years ago
7 0

<u>We are given:</u>

P1 = 3 atm                  T1 = 623 K <em>(350 + 273)</em>

P2 = x atm                 T2 = 523 K <em>(250 + 273)</em>

<em />

<u>Solving for x:</u>

From the idea gas equation:

PV = nRT

since number of moles (n) , Volume (V) and the Universal Gas constant(R) are constants;

P / T = k   (where k is a constant)

the value of  k will be the same for a gas with variable pressure and temperature and constant moles and volume

Hence, we can say that:

P1 / T1 = P2 / T2

3 / 623 = x / 523

x = 523 * 3 / 623

x = 2.5 atm (approx)

Therefore, the final pressure is 2.5 atm

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tia_tia [17]

Hello!

The answer is C. Hibernate during the cold winter months.

Why?

Alpine marmots are known for having a long hibernation duration which starts in October (winther) and ends in April (summer) (about 7 months). During this long period, they are able to reduce their bear beats from 200 per minute to just 30 or 38 beats, and their breaths from 60 breaths/minute to 1-3 breaths/minute, guaranteeing an extreme energy saving process.

Have a nice day!

5 0
2 years ago
Why did mendeleev leave blank spaces in his version of the periodic table?.
Leviafan [203]
<h2>Answer: He left blank spaces for <em>undiscovered</em> elements to be added to the table.</h2>

Explanation:

3 0
1 year ago
Read 2 more answers
5. How can you tell the difference between CuS and Cu2S
puteri [66]

Answer:

Density, melting point. and magnetic properties  

Explanation:

I can think of three ways.

1. Density

The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.

It should be possible to distinguish these even with high school equipment.

2. Melting point

Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).

A Bunsen burner can easily reach these temperatures.

3. Magnetic properties

You can use a Gouy balance to measure the magnetic susceptibilities.

In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.

In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.

A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.

A sample of CuS will be attracted by the magnetic field and show an increase in weight.

In the picture below, you can see the sample partially suspended between the poles of an electromagnet.

 

6 0
3 years ago
Which coefficient before potassium (K) balances this chemical equation?
miskamm [114]

Answer:

The coefficient before potassium (K) balances this chemical equation is 2.

Explanation:

_K +Cl₂ → 2KCl

K =1 ; Cl =2

K=1 × 2 = 2

Cl = 1 × 2 = 2

2 K +Cl₂ = 2 KCl

5 0
3 years ago
Read 2 more answers
Question 3) A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of
Masja [62]

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

<u>Explanation:</u>

We have the chemical equation,

HF (aq)+NaOH(aq)->NaF(aq)+H2O

To find how many moles have been used in this

c= n/V=> n= c.V

nHF=0.250 M⋅1.5 L=0.375 moles HF

Simillarly

nF=0.250 M⋅1.5 L=0.375 moles F

nHF=0.375 moles - 0.250 moles=0.125 moles

nF=0.375 moles+0.250 moles=0.625 moles

[HF]=0.125 moles/1.5 L=0.0834 M

[F−]=0.625 moles/1.5 L=0.4167 M

To determine the problem using the Henderson - Hasselbalch equation

pH=pKa+log ([conjugate base/[weak acid])

Find the value of Ka

pKa=−log(Ka)

pH=−log(Ka) +log([F−]/[HF]

pH= -log(3.5 x 10 ^4)+log(0.4167 M/0.0834 M)

pH=-log(3.5 x 10 ^4)+log(4.996)

pH= -4.54+0.698

pH=-(-3.84)

pH=3.84

The pH of the solution after adding 0.150 moles of solid LiF is 3.84

5 0
3 years ago
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