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mixas84 [53]
2 years ago
11

What if there was an inverse relationship between the temperature and the volume? How do you think the world might be different?

How would this change our day-to-day lives?
Chemistry
1 answer:
kari74 [83]2 years ago
5 0

If there was an inverse relationship between the temperature and the volume, our daily lives change because in high temperature things will contract.

<h3>What if there was an inverse relationship between the temperature and the volume?</h3>

If there was an inverse relationship between the temperature and the volume then with increasing temperature decrease occur in the volume of a substance. If this type of relationship is present in the world, the objects will contract when the temperature is high and expand when the temperature is low which make the solid materials expand at winter and contract at summer season.

So we can conclude that if there was an inverse relationship between the temperature and the volume, our daily lives change because in high temperature things will contract.

Learn more about temperature here: brainly.com/question/25677592

#SPJ1

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An OH group attached to a hydrocarbon is called a _________ group whereas ______________ is a polyatomic ion with a charge of __
rodikova [14]

Answer:

sijshsjdjdjdjdjakskskkskzjzz

3 0
3 years ago
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
3 years ago
PLZ HELP!!
Nataly [62]

A and B are experiencing winter. The picture which isn't available here in this question is attached below.

Option C.

<h3><u>Explanation:</u></h3>

The earth is tilt by an angle of 23.2° to the vertical plane. This makes the seasonal variation of earth, because in some time of the year, the northern hemisphere faces the sun directly, experiencing summer and then southern hemisphere is away from summer experiencing winter and vice versa. The summer occurs when the place directly faces the sun. And the winter happens when the place obliquely faces the sun or doesn't face the sun at all.

Here in this diagram, we can see that the points A and B are the north pole and the part in northern hemisphere respectively which aren't facing the sun directly, whereas C and D are facing the sun. Thus the southern hemisphere is experiencing summer and the northern hemisphere the winter.

4 0
3 years ago
Calculate the pOH of a 0.0143 M NaOH solution at 25 ⁰C.
arsen [322]

Answer:

A.) 1.845

Explanation:

You can find the pOH using the following equation:

pOH = -log[OH⁻]

Since NaOH dissociates into 1 Na⁺ and 1 OH⁻, the concentration of both ions is 0.0143 M.

pOH = -log[OH⁻]

pOH = -log[0.0143]

pOH = 1.845

5 0
2 years ago
This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press
Solnce55 [7]

Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

           [A]₀ is the initial concentration of A

           k is the rate constant, and

           t is the time

We also know that for a first order reaction

           k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 /  35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀  = - kt

thus:

(pPH₃ )t  = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

3 0
3 years ago
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