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kati45 [8]
2 years ago
8

. If measurements of a gas are 75L and 300 kilopascals and then the gas is measured a second time and found to be 50L, describe

what had to happen to the pressure (if temperature remained constant). Include which law supports this observation.
Physics
2 answers:
mash [69]2 years ago
8 0
<h2>Answer:</h2>

<u>The pressure increased to 450 Kilo pascals.</u>

<h3>Explanation:</h3>

This the condition of Boyle's law:

P1V1 = P2V2

By putting values in above formula:

= 300 * 75 = P2 * 50

= P = 300 * 75/50

= p = 450 Kilo pascals

In this question description, two variables volume and pressure are involved while the temperature remains constant. So it is the condition of <em><u>Boyle's law which states that the Pressure and volume of a gas are inversely proportional if the temperature of the gas remains constant.</u></em>

julia-pushkina [17]2 years ago
3 0
By Boyle's law:

P₁V₁ = P₂V₂

300*75 = P<span>₂*50

</span>P<span>₂*50= 300*75
</span>
P<span>₂ = 300*75/50 = 450
</span>
P<span>₂ = 450 kiloPascals.

The pressure has increased as a result of compression of gas.

Boyle's Law supports this observation.</span>
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1) The equilibrium constant Kc for the reaction N 2(g) + O 2(g) 2NO(g) at 1200 C is 1.00x 10^-5. Calculate the molar concentrati
Elina [12.6K]

Explanation:

1) N₂ + O₂ → 2 NO

Kc = [NO]² / ([N₂] [O₂])

Set up an ICE table:

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\N_{2}&0.114&-x&0.114-x\\O_{2}&0.114&-x&0.114-x\\NO&0&+2x&2x\end{array}\right]

Plug into the equilibrium equation and solve for x.

1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))

1.00×10⁻⁵ = (2x)² / (0.114 − x)²

√(1.00×10⁻⁵) = 2x / (0.114 − x)

0.00316 = 2x / (0.114 − x)

0.00361 − 0.00316x = 2x

0.00361 = 2.00316x

x = 0.00018

The volume is 1.00 L, so the concentrations at equilibrium are:

[N₂] = 0.114 − x = 0.11382

[O₂] = 0.114 − x = 0.11382

[NO] = 2x = 0.00036

2(a) Cl₂ → 2 Cl

Kc = [Cl]² / [Cl₂]

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\Cl_{2}&2.0&-x&2.0-x\\Cl&0&+2x&2x\end{array}\right]

1.2×10⁻⁷ = (2x)² / (2 − x)

1.2×10⁻⁷ (2 − x) = 4x²

2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²

2.4×10⁻⁷ ≈ 4x²

x² ≈ 6×10⁻⁸

x ≈ 0.000245

2x ≈ 0.00049

2(b) F₂ → 2 F

Kc = [F]² / [F₂]

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\F_{2}&2.0&-x&2.0-x\\F&0&+2x&2x\end{array}\right]

1.2×10⁻⁴ = (2x)² / (2 − x)

1.2×10⁻⁴ (2 − x) = 4x²

2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²

2.4×10⁻⁴ ≈ 4x²

x² ≈ 6×10⁻⁵

x ≈ 0.00775

2x ≈ 0.0155

F₂ dissociates more, so Cl₂ is more stable at 1000 K.

7 0
3 years ago
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of
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Answer:

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We have to find the change in the ball's momentum. It is given by :

\Delta p=p_f-p_i

\Delta p=m(v-u)

\Delta p=0.275\ kg(1.7\ m/s-(-2.4\ m/s))

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The error percentage in measuring the volume of the cylindrical box is 6.3 x 10⁻³ %

<h3>Error percentage in measuring the volume of the cylinder</h3>

The total error percentage in measuring the volume of the cylinder is calculated as follows;

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Error in measuring the volume = 6.3 x 10⁻⁵ x 100% = 6.3 x 10⁻³ %

Learn more about percentage error here: brainly.com/question/5493941

#SPJ1

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