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elena-14-01-66 [18.8K]
3 years ago
10

An object starts at position 12 on a horizontal line with a reference point of o. What is the position of the object if it moves

14
units to the left?
0-26
O-2
OOO
Chemistry
1 answer:
eimsori [14]3 years ago
6 0

Answer:

-2

Explanation:

Consider object is starting 12 units right from the reference point which is 0.

Assign the right direction positive sign.

when object is moving 14 units on left direction the position of object will be two units to the left side of reference point.

Assign the left direction negative sign position will be -2.

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Find δs∘ for the reaction between nitrogen gas and hydrogen gas to form ammonia:12n2(g) 32h2(g)→nh3(g)
Dima020 [189]

Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).

What is standard entropy?

The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."

Calculation:

Balancing the given reaction following-

1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)

ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]

Here S° = standard entropy of the system

Insert into the aforementioned equation all the typical entropy values found in the literature:

ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]

⇒ΔS° = - 99.4 J/K

Therefore, the standard entropy, ΔS° is -99.4 J/K.

Learn more about standard entropy here:

brainly.com/question/14356933

#SPJ4

5 0
1 year ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
Calculate the equilibrium constant K for the following reaction: H2(g) +
SIZIF [17.4K]

Answer:

192.9

Explanation:

From the question,

Ke = [HCL]²/[H₂][CL₂].......................... Equation 1

Where Ke = Equilibrium constant.

Given: [HCL] = 0.0625 M, [H₂] = 0.0045 M, [CL₂] = 0.0045 M

Substitute these values into equation 1

Ke = (0.0625)²/(0.0045)(0.0045)

ke = (3.90625×10⁻³)/(2.025×10⁻⁵)

ke = 1.929×10²

ke = 192.9

Hence the equilibrium constant of the system = 192.9

5 0
3 years ago
If faucet drips 5 mL of water each minute, what is the volume of water dripped at the end of five minutes?
kiruha [24]
If the faucet drops at 5 mL per minute you just have to do the following steps to find out.

5 mL for the dripping rate times 5 for the minutes.

The answer would result being 25mL

Hope this helps! Stay safe!
6 0
3 years ago
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NISA [10]

Answer:

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Explanation:

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3 years ago
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