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lesantik [10]
3 years ago
15

The force needed to overcome kinetic energy friction is usually less than that needed to overcome static friction

Chemistry
1 answer:
Anestetic [448]3 years ago
6 0

the answer is true becuase The force needed to overcome static friction is usually less than that needed to overcome ______ friction

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Which of the following provides the most evidence for the theory of plate tectonics? A. the spreading of the sea floor B. the co
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Answer: the correct answer is a.

Explanation: a spreading of the sea floor is the correct answer.

3 0
3 years ago
Consider the decomposition of the compound C5H6O3 as follows: C5H6O3(g)  C2H6(g) + 3CO(g) A 5.63 g sample of pure C5H6O3(g) was
deff fn [24]

Answer:

K_{eq}=1.02x10^{-4}

Explanation:

Hello,

In this case, the first step is to compute the initial moles of C₅H₆O₃ as shown below:

5.63gC_5H_5O_3*\frac{1molC_5H_5O_3}{114gC_5H_5O_3}=0.0494molC_5H_5O_3

After that, by knowing that the final pressure is 1.63 atm, one computes the total moles at the equilibrium as follows:

n_{total}^{eq}=\frac{P_{total}^{eq}V}{RT}=\frac{1.63atm*2.50L}{0.082\frac{atm*L}{mol*K}*473.15K} =0.105mol

Then, by knowing the moles at the equilibrium considering the change "x", which yields to:

\ \ \ \ \  C_5H_6O_3(g) \leftrightarrow  C_2H_6(g) + 3CO(g)\\I\ \ \ \ \ 0.0494mol\ \ \ \ \ \ 0mol \ \ \ \ \ \ \ \ 0mol\\C\ \ \ \ \ \ -x\ \ \ \ \ \ \ \ \ \ \  \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x\\E\ \ 0.0494mol-x\ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ 3x

The total moles at the equilibrium turn out:

n_{total}^{eq}=0.0494mol-x+x+3x

By solving for "x", we've got:

3x=0.105mol-0.0494mol\\x=\frac{0.0556mol}{3}\\x=0.0185mol

Finally, the equilibrium constant is:

K_{eq}=\frac{(x)(3x)^3}{0.0494-x}=\frac{(0.0185mol)(3*0.0185mol)^3}{0.0494mol-0.0185mol}=1.02x10^{-4}

Best regards.

4 0
3 years ago
At constant temperature, 534 mL of a gas at 894 torr is compressed to 167 mL What is the new pressure in atmospheres?
AURORKA [14]

Answer:

3.762 atm.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

  • If n and T are constant, and have different values of P and V:

<em>(P₁V₁) = (P₂V₂)</em>

<em></em>

  • Knowing that:

P₁ = 894.0 torr, V₁ = 534.0 mL,

P₂ = ??? torr, V₂ = 167.0 mL.

  • Applying in the above equation

(P₁V₁) = (P₂V₂)

<em>∴ P₂ = (P₁V₁)/V₂</em> = (894.0 torr)(534.0 mL)/(167.0 mL) = <em>2859 torr.</em>

  • To convert from torr to atm:

1.0 atm = 760.0 torr.

<em>∴ P₂ </em>= (2859 torr)(1.0 atm/760 torr) = <em>3.762 atm.</em>

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