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Archy [21]
4 years ago
11

How much work is required to lift a 0.500 kg block 18.5 m?

Chemistry
1 answer:
kipiarov [429]4 years ago
3 0

Answer:

Work = 90.65 j

Explanation:

Given data:

Mass = 0.500 Kg

Distance = 18.5 m

Work done = ?

Solution:

Work = force . distance

Force = mg

Work = mg.distance

Work = mgh

Work = 0.500 Kg × 9.8 m/s²× 18.5 m

Work = 90.65 Kg .m²/s²

Kg .m²/s² = j

Work = 90.65 j

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Titanium and chlorine react to form titanium(IV) chloride, like this:
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Answer:

the value of equilibrium constant for the reaction is 8.5 * 10⁷

Explanation:

Ti(s) + 2 Cl₂(g) ⇄ TiCl₄(l)

equilibrium constant Kc = \frac{1}{[Cl_2]^2}

Given that,

We are given:

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We calculate the No of mole = mass / molar mass

mass of chlorine gas = 1.67 g

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= 7.0L

Concentration of chlorine is = no of mole / volume

= 0.024 / 7

= 3.43 * 10⁻³M

equilibrium constant Kc  = \frac{1}{[Cl_2]^2}

= \frac{1}{[3.43 * 10^-^3]^2}

= 8.5 * 10⁷

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Lipids are soluble in a class of solvents called????
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