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3 years ago

The diagram below shows the temperature dropping from 80°C to 20°C.

Chemistry

2 answers:

Neko [114]3 years ago

8 0

Answer:

Kinetic energy decreases as temperature decreases.

Explanation:

From the description that the system at 80°C has longer arrows, or move faster than the system at 20°C, having shorter actors indicating a slower motion, we can conclude that the kinetic energy of a body depends on its temperature.

If the system at 80°C shows a greater kinetic energy (faster motion of particles) than the system at 20°C, it then implies that decreasing the temperature of the body decreases its kinetic energy.

dexar [7]3 years ago

3 0

Answer:

C on edge

Explanation:

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When balancing a chemical equation, the number of h atoms in 2 ch4 is eight?

Viktor [21]

Yes... that is correct.

CH4 is methane so the coefficent in front of it would double the number of atoms of each element

6 0

3 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

3 years ago

Scientists use what properties to help them identify and classify matter

Volgvan

Scientists use the physical and chemical properties to help them identify and classify matter. These physical and chemical properties are in a macro-perspective, in which these matter contains compounds, elements and atoms. Hence, matter can be classified in various ways, 
1. Atomic number either atomic mass each element has
2. By substance of that matter either pure substance or mixed substance 
3. If they cannot reduce a certain substance into a much smaller quantified atomic structure then they they’ll use (2) to identify and classify it.

5 0

3 years ago

How many grams of lithium sulfide must be dissolved in 1600.0 g water to make a 2.0 molal solution?

klasskru [66]

Molality is the number of moles of solutes in 1 kg of solvent.
the molality of solution to be prepared is 2.0 molal.
therefore 2 moles in 1 kg water.
the mass of Li₂S required is - 46 g/mol x 2.0 mol = 92 g
the mass in 1 kg of solvent is - 92 g
Therefore mass of Li₂S required in 1600.0 g is - 92 g/kg x 1.6 kg = 147.2 g

7 0

2 years ago

PLEASE HELP!!!

Inga [223]

Answer:

$d=1.5\ g/cm^3$