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Delicious77 [7]
2 years ago
15

In which way would a drought affect how rodents get their energy?

Chemistry
1 answer:
love history [14]2 years ago
3 0

Answer:

<h3>When a drought occurs, their food supply can shrink and their habitat can be damaged. ... Losses or destruction of fish and wildlife habitat. Lack of food and drinking water for wild animals. Increase in disease in wild animals, because of reduced food and water supplies.</h3><h3>While insects and cacti might provide a meagre supply of water, most desert animals survive by being what Price calls "water misers". ... To perform this feat, they have evolved specialized kidneys with extra microscopic tubules for extracting water from urine.</h3>

Explanation:

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The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0
3 years ago
Name the following compound:<br> CH3-CH2-CH2-CH2-CH3<br> CH3 CH3
stepan [7]

Answer:

<u><em>Pentane </em></u>

Explanation:

since we have in here CH3-CH2-CH2-CH2-CH3 5 Carbon atoms and 12 Hydrogen making it C_{5} H_{12}

6 0
3 years ago
In the first 85.0 s of this reaction, the concentration of no dropped from 1.12 m to 0.520 m . calculate the average rate of the
goldenfox [79]

In the first 85.0 s of this reaction, the concentration of no dropped from 1.12 m to 0.520 m .

What is rate of a reaction?

The speed at which a chemical reaction takes place is the rate of the reaction. It is the concentration change per unit time of a reactant in a reaction.

Since the concentration of NO reduces to half its initial concentration in 85 seconds that is from 1.12m to 0.520m, it can be said that 85 seconds is the half life interval for the reaction, <u>Hence on average, </u><u>half reaction</u><u> is completed in the time interval of </u><u>85 seconds</u><u>.</u>

To learn more about rate of a reaction from the given link below,

brainly.com/question/12172706

#SPJ4

7 0
1 year ago
Someone help me rn! ASAP
Olenka [21]

Answer:thermal energy

Explanation:Thermal radiation is the emission of electromagnetic rays from all matter that’s greater then zero

3 0
3 years ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
Likurg_2 [28]

Answer: 281 hours

Explanation:-

1 electron carry charge=1.6\times 10^{-19}C

1 mole of electrons contain=6.023\times 10^{23} electrons

Thus  1 mole of electrons carry charge=\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

96500\times 2=193000Coloumb of electricity deposits 1 mole or 63.5 g of copper

0.0635 kg of copper is deposited by 193000 Coloumb

11.5 kg of copper is deposited by=\frac{193000}{0.0635}\times 11.5=34952756 Coloumb

Q=I\times t

where Q= quantity of electricity in coloumbs  = 34952756 C

I = current in amperes = 34.5 A

t= time in seconds = ?

34952756 C=34.5A\times t

t=1013123sec=281hours

Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.

8 0
3 years ago
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