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Sindrei [870]
3 years ago
12

At equilibrium, ________. At equilibrium, ________. all chemical reactions have ceased the rate constants of the forward and rev

erse reactions are equal the rates of the forward and reverse reactions are equal the value of the equilibrium constant is 1
Chemistry
1 answer:
balandron [24]3 years ago
6 0

Answer:

<em>At equilibrium, the rate of the forward, and the reverse reactions are equal.</em>

Explanation:

In an equilibrium chemical reaction, the rate of forward reaction, is equal to the rate of reverse reaction. Note that the reactions does not cease at equilibrium, but rather, the reactants are converted to product, at the same rate at which the product is also being converted into the reactants in the reaction. When chemical equilibrium is reached, a careful calculation of the value of equilibrium constant is approximately equal to 1.

NB: If the value of equilibrium constant is far far greater than 1, then the reaction will favors more of the forward reaction, and if far far less than 1, the reaction will favor more of the reverse reaction.

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Answer:

(d)

Explanation:

Carbonyl group can be the placement of kerosene sugar

6 0
3 years ago
the equilibrium constant K for a certain reversible reaction is 1.32 × 10−4. Which of the following is true for the reaction? Th
Mazyrski [523]

<u>Answer:</u>

<em>The reverse reaction is more favourable than the forward reaction. </em>

<u>Explanation</u>:

<em>The value of equilibrium constant in this case is  1.32 \times 10 ^-^4 (0.000132)</em>.  A high value of equilibrium constant indicates that the forward reaction is more favourable.  

Thus the<em> resultant of the reaction will have higher concentration of products.  </em>When the value of equilibrium constant is less the reverse reaction will be more favoured than the forward reaction.

Thus the resultant of the reaction will have more reactants than products. <em>Here equilibrium constant has a very low value which favours reverse reaction. </em>

7 0
3 years ago
What is the theoretical mass of CO2 + H2O lost per gram of NaHCO3?
Mama L [17]

Answer:

The theoretical mass of CO2 = 2.09 grams

The theoretical mass of H2O =  0.429 grams

The theoretical  mass of CO2+ H2O = 2.519 grams

Explanation:

Step 1: Data given

Mass of NaHCO3 = 1 gram

Molar mass NaHCO3 = 84.00 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

4NaHCO3 + O2 → 4CO2 + 2H2O + 2Na2O2

Step 3: Calculate moles of NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

Moles NaHCO3 = 1.0 gram / 84.00 g/mol

Moles NaHCO3 = 0.0119 moles

Step 4: Calculate moles CO2 and H2O

For 4 moles NaHCO3 we need 1 mol O2 to produce 4 moles CO2 and 2 moles H2O and 2 moles Na2O2

For 0.0119 moles NaHCO3 we'll have 4*0.0119= 0.0476 moles CO2

and 2* 0.0119 = 0.0238 moles H2O

Step 5: Calculate mass CO2 and H2O

Mass = moles * molar mass

Mass CO2 = 0.0476 moles * 44.01 g/mol

Mass CO2 = 2.09 grams

Mass H2O = 0.0238 moles ¨18.02 g/mol

Mass H2O = 0.429 grams

The theoretical mass of CO2 = 2.09 grams

The theoretical mass of H2O =  0.429 grams

The theoretical  mass of CO2+ H2O = 2.519 grams

6 0
4 years ago
Based on this graph what was the oxygen content of the earths atmosphere about 500million years ago
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12 percent is your answer :))

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Explanation:

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