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Kitty [74]
3 years ago
11

A group of scientists are studying the effects of different fertilizers on the growth of pea plants. They began their study with

10 pea plants. They gave each plant a different type of fertilizer and tracked their growth over a period of 2 months. The results of this experiment weren’t considered valid by other scientists because there was no control group. Which of the following represents an appropriate control group for this experiment? A. A group of plants that’s subjected to two fertilizers B. A group of plants that’s subjected to one fertilizer C. A group of plants that’s not subjected to any fertilizer D. A group of plants that’s subjected to all types of fertilizers
Chemistry
2 answers:
Ymorist [56]3 years ago
5 0
Probably A hope this helps.
docker41 [41]3 years ago
3 0
From what i read I’d say the answer is B
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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
PLEASE HELP!!!
Elena L [17]

Answer:

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2 years ago
Once formed, how are coordinate covalent bonds different from other covalent bonds?
-Dominant- [34]

Answer:

\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}

Explanation:

A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).

Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.

Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.

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3 years ago
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Eddi Din [679]
The answer is (3) a homogeneous mixture. The difference between homogeneous and heterogeneous mixture is the degree of the mixture being mixed. Due to the completely dissolved and the dissolving ability of KCl, we can get the answer,
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