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1 month ago

compare and contrast melting 10 kg of ice freezing 1 kg of water. be sure to address temperature,heat flow, and thermal energy

1 answer:

6 0

<span>Melting is a process that absorbs heat (endothermic), while freezing is one that releases heat (exothermic), or it can be wrongly perceived in thermodynamic terms as "absorbing cold." The standard enthalpy of fusion for water applies to both processes, but remember standard enthalpy is measured in energy per mass, so melting 10 kg of ice will require 10 times the energy compared to freezing 1 kg of water. For both cases, assuming the water is pure and the pressure remains at standard atmospheric levels, the temperature of the solid and the liquid will stabilize at exactly 0 degrees Celsius (273 K) during the entire phase transition.</span>

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An experimental drug, D, is known to decompose in the blood stream. Tripling the concentration of the drug increases the decompo

lions [2927]

Answer:

The rate law for the decomposition reaction is:

$R=k[D]^2$

The unit for the rate constant will be $M^{-1}s^{-1}$

Explanation:

$D\rightarrow Product$

The rate law can be expressed as:

$R=k[D]^x$ ..[1]

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

$[D]'=3[D]$

$R'=9\times R$

$R'=k[D]'^x$ ...[2]

[1] ÷ [2]

$\frac{R}{R'}=\frac{k[D]^x}{k[D']^x}$

$\frac{R}{9R}=\frac{k[D]^x}{k[3D]^x}$

$9=3^x$

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

$R=k[D]^2$

The unit for the rate constant will be:

$k=\frac{R}{[D]^2}=\frac{M/s}{(M)^2}=M^{-1}s^{-1}$

The unit for the rate constant will be $M^{-1}s^{-1}$ .

5 0

2 months ago

Hydrogen has three isotopes with mass numbers of 1, 2, and 3 and has an average atomic mass of 1.00794 amu. This information ind

VMariaS [2998]

The answer is actually 3, believe me.

Explanation:

7 0

3 months ago

Read 2 more answers

The volume of the ideal gas in the container is 9 m3. What would the volume be if four more blocks were placed on top of the lid

castortr0y [3046]

Response:
4.5 m³

Resolution:
The statement indicates the presence of two blocks on a lid of a container with a volume of 9 m³. The lid's weight is equal to that of the two blocks. Thus, there were initially four blocks (or 4 atm pressure) acting on a volume of 9 m³.

After adding four additional blocks on the lid, the pressure rises from 4 atm to 8 atm (2 atm from the lid, 2 atm from the original blocks, and 4 atm from the new blocks).

Hence, The data established is,

P₁ = 4 atm

V₁ = 9 m³

P₂ = 8 atm

V₂ =?

Using Boyle's Law,

P₁ V₁ = P₂ V₂

Resolving for V₂,
V₂ = P₁ V₁ / P₂

Substituting values yields:
V₂ = (4 atm × 9 m³) ÷ 8 atm

V₂ = 4.5 m³

0 0

2 months ago