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nika2105 [10]
3 years ago
12

Help lol sorry if I’m spamming this is just really hard for me

Chemistry
2 answers:
yanalaym [24]3 years ago
7 0
True!! Enwnwnwnwna. Anwnwnwnwwnan
galben [10]3 years ago
4 0
It’s true:)
The subscript on chemical equations tells you how many atoms the elements have. If there is no subscript then there is only 1 atom.
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The molar mass of Cr(OH)2 is:
son4ous [18]

The molar mass of Cr(OH)_2 is 86.02 g/mole .

<h3><u> Explanation:</u> </h3>

The molar mass of a chemical compound is represented as the mass of a unit of that compound separated by the number of substances in that unit, measured in moles. The molar mass is a volume, not molecular, the property of a substance.

The molar mass is a percentage of various examples of the compound, which usually change in mass due to the appearance of isotopes.

From the below attached table, the Molar mass of Cr(OH)_2 is 86.0108 g/mol.

3 0
3 years ago
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
Consider the following reactions. (Note: (s) = solid, (l) = liquid, and (g) = gas.) ½H2(g) + ½I2(g) → HI(g), ΔH = +6.2 kcal/mole
BARSIC [14]
½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS
4 0
3 years ago
Read 2 more answers
Select the person responsible for the following: printing press
Alinara [238K]
That would be <span>Johannes Gutenberg a German inventor from the 1400's </span>
4 0
3 years ago
What name is given to the minimum amount of energy needed by particles for them to react when they collide?
makkiz [27]

Answer:

Activation

Explanation:

8 0
3 years ago
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