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3 years ago

I hope someone can help me here please don't answer nonsense!

Chemistry

1 answer:

viva [34]3 years ago

6 0

The final pressure of the balloon with no temperature change is 4,560 torr.

<h3>Final pressure of the gas</h3>

The final pressure of the gas is determined by applying Boyle's law as shown below.

P₁V₁ = P₂V₂

P₂ = (P₁V₁)/(V₂)

P₂ = (1.5 x 4)/(1)

P₂ = 6 atm

1 atm = 760 torr

6 atm = ?

= 4,560 torr

Thus, the final pressure of the balloon with no temperature change is 4,560 torr.

Learn more about final pressure here: brainly.com/question/25736513

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How do u find the volume of a sphere

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Answer:

$V=\frac{4}{3} \pi r^3$

Explanation:

plug in the radius for r and the v doesn't matter

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3 years ago

At 500K, the equilibrium constant for the reaction N2O4D2NO2 is 1.5 x 10^3. What is the equilibrium constant for the reaction: 6

Veronika [31]

Answer:

K = 2.96x10⁻¹⁰

Explanation:

Based on the initial reaction:

N2O4 ⇄ 2NO2; K = 1.5x10³

Using Hess's law, we can multiply this reaction changing K:

3 times this reaction:

3N2O4 ⇄ 6NO2; K = (1.5x10³)³ =3.375x10⁹

The inverse reaction has a K of:

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3 years ago

Question 23 (5 points)

iren [92.7K]

Answer:

λ = 6.25 × 10⁻¹⁰ m

x-ray

General Formulas and Concepts:

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c = λν

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<u>Step 1: Define</u>

C = 3.0 × 10⁸ m/s

ν = 4.80 × 10¹⁷ Hz (s⁻¹)

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3.0 × 10⁸ m/s = λ(4.80 × 10¹⁷ Hz)

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5 0

3 years ago

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

ira [324]

Explanation:

The given data is as follows.

m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)

Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.

Also, we know that the relation between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

$\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= -554.12 J/K + 647.49 J/K

= 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

3 years ago

How many moles are there in 3.01 x 1024 atoms of argon?

Ahat [919]

Answer:

Try this for answer 3082.24

6 0

3 years ago