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melomori [17]
2 years ago
8

250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze?

Chemistry
1 answer:
Stolb23 [73]2 years ago
5 0

250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze: Yes, because there is enough energy.

<h3>At what temperature would a sample of water freeze?</h3>
  • Note from the Facilitator: At certain temperatures, water changes its condition due to temperature variations. At sea level, fresh water changes from a solid to a liquid at 32°F (0°C). Liquid water freezes at temperatures below 32°F (0°C); this temperature is known as the freezing point of water.
  • The fact that a single water molecule cannot transform into a solid, liquid, or gas is the answer. These names refer to collective behaviors of water molecules rather than to individual molecules.
  • For instance, the solid (ice) has a collection of molecules that are bound together and arranged in a predictable manner. That cannot be accomplished by a single molecule alone

250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze: Yes, because there is enough energy.

To learn more about water freezing, refer to:

brainly.com/question/15209660

#SPJ9

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For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The
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Answer : The rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-2}M^{-2}s^{-1}

Thus, the value of the rate constant 'k' for this reaction is 7.5\times 10^{-2}M^{-2}s^{-1}

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1

\text{Rate}=4.25\times 10^{-2}Ms^{-1}

Therefore, the rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

3 0
3 years ago
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