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Dahasolnce [82]
3 years ago
5

in your notebook,write the appropriate tempo for the following actions.Then demostrate the actions then,demostrate the action in

front of your parents,guardian
Chemistry
1 answer:
pogonyaev3 years ago
6 0

Answer:

cell :—

•They lack a well-defined nucleus, have a nucleoid instead.

•Usually have double-stranded, circular DNA.

•Do not have mitochondria.

Eukaryotic cell :—

•Have a well-defined nucleus enclosed in the nuclear membrane.

•Have a linear double-stranded nucleus.

•Mitochondria are present.

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The balanced equation for the reaction of ammonia and oxygen is the following. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) The stand
ale4655 [162]

Answer:

ΔS° = 180.5 J/mol.K

Explanation:

Let's consider the following reaction.

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.

ΔS° = ∑np × S°p - ∑nr × S°r

where,

ni are the moles of reactants and products

S°i are the standard molar entropies of reactants and products

ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))

ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol

ΔS° = 180.5 J/K

This is the change in the entropy per mole of reaction.

7 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
Be sure to answer all parts. Carbon dioxide (CO2) is the gas that is mainly responsible for global warming (the greenhouse effec
Elena L [17]

Answer:

1.60x10⁶ billions of g of CO₂

Explanation:

Let's calculate the production of CO₂ by a single human in a day. The molar mass of glucose is 180.156 g/mol and CO₂ is 44.01 g/mol. By the stoichiometry of the reaction:

1 mol of C₆H₁₂O₆ -------------------------- 6 moles of CO₂

Transforming for mass multiplying the number of moles by the molar mass:

180.156 g of C₆H₁₂O₆ ----------------- 264.06 g of CO₂

4.59x10² g ---------------- x

By a simple direct three rule:

180.156x = 121203.54

x = 672.77 g of CO₂ per day per human

So, in a year, 6.50 billion of human produce:

672.77 * 365 * 6.50 billion = 1.60x10⁶ billions of g of CO₂

5 0
3 years ago
Under what conditions does the molar volume of a gas decrease
nydimaria [60]

Answer:

Gases are compressible, meaning that when put under high pressure, the particles are forced closer to one another. This decreases the amount of empty space and reduces the volume of the gas. Gas volume is also affected by temperature. When a gas is heated, its molecules move faster and the gas expands.

4 0
3 years ago
Read 2 more answers
PLEASE HELP Which of the answers below would be a correct chemical symbol of an element?
UNO [17]

Answer:

Hg

Explanation:

8 0
3 years ago
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