Question 4 of 10

Chemistry

1 answer:

fredd [130]3 years ago

6 0

Answer:

Explanation:

because of the law of gravity

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The line graph shows the heights of plants grown in fertilized and unfertilized soil. Based on this information, what will most

Helen [10]

Answer:

g) Neither plant should increase by 1 cm in height.

Explanation:

See the graph for this question on the figure attached.

The growing of the plant A is represented by the line that goes above the other. At start, that line has a slope that rises about 0.75 cm ( height increase) in 1 day. From the day 2 and forward the slope of the line decreases. The line reaches its highest point about at day 4 and seems to start decreasing. Thus, you should predict that on the day six it most likely does not increase in height.

The growing of the plant B is represented by the line drawn below the other. As for the plant B, the growing decreases with the number of days. Between the days 4 and 5 the line is almost flat, which means that most likely this plant will not grow on the day six or grow less than 0.5 cm.

Thus, for both plants you can say that on day six, most likley, neither should increase by 1 cm in height (option g).

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3 years ago

What type of substance is dirt?!

stealth61 [152]

Answer:

A substance containing only one type of atom or one type of molecule is a pure substance. Most of the matter around us, however, consists of mixtures of pure substances. Air, wood, rocks and dirt are examples of such mixtures. Mixtures can be further classified as Homogeneous and Heterogeneous.

4 0

3 years ago

Read 2 more answers

The standard free energy of activation of a reaction A is 81.9 kJ mol–1 (19.6 kcal mol–1) at 298 K. Reaction B is one million ti

klasskru [66]

Answer:

54.9 kJ/mol

Explanation:

The relation between the activation energy (Ea) and the rate constant (k) is given by the Arrhenius equation.

$k=A.e^{-Ea/RT}$

where,

A is a collision factor

R is the ideal gas constant

T is the absolute temperature

Reaction B is one million times faster than reaction A at the same temperature. So $k_{B}=10^{6} k_{A}$ .

Then,

$k_{B}=10^{6} k_{A}\\A.e^{-Ea_{B}/RT}=10^{6}A.e^{-Ea_{A}/RT}\\e^{-Ea_{B}/RT}=10^{6}e^{-Ea_{A}/RT}\\ln(e^{-Ea_{B}/RT})=ln(10^{6}e^{-Ea_{A}/RT})\\\frac{-Ea_{B}}{RT} =ln10^{6} -\frac{Ea_{A}}{RT} \\Ea_{B}=(ln10^{6} -\frac{Ea_{A}}{RT}).(-RT)=(ln10^{6}-\frac{89.1kJ/mol}{(8.314\times 10^{-3} kJ/mol.K).298K} ).(-8.314\times 10^{-3} \frac{kJ}{mol.K}.298K )=54.9kJ/mol$