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son4ous [18]
3 years ago
9

In the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not

affect the reaction rate. What is the rate law for this reaction?
A) rate = k[B]
B)rate = k[A]2
C)rate = k[A][B]
D)rate = k[A]
Chemistry
1 answer:
Kitty [74]3 years ago
6 0
Here, the rate of reaction would be: rate = k[A]

In short, Your Answer would be Option D

Hope this helps!
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2 bromocyclopentanamine structural formula
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Answer:

The structure is given in attached file.

Explanation:

Explanation

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5 0
3 years ago
Given the equation representing a system at equilibrium:
polet [3.4K]
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3. adding a noble gas
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hope this helps
3 0
3 years ago
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Blizzard [7]

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Volume

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3 0
3 years ago
Read 2 more answers
CaCO3(s) ⇄ CaO(s) + CO2(g) 0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K.
Montano1993 [528]

Answer:

The final mass of CaCO3 is 10.68 grams

Explanation:

Step 1: Data given

Number of moles CaCO3 = 0.100 moles

Number of moles CaO = 0.100 moles

Volume = 10.0 L

When equilibrium is reached the pressure of CO2 is 0.220 atm. 0.250 atm of CO2 is added, while keeping the temperature constant

Step 2: The balanced equation

CaCO3(s) <==> CaO(s) + CO2(g)

Step 3: Calculate moles of CO2

n = PV/RT

⇒n = the initial number of moles CO2 = TO BE DETERMINED

⇒P = the pressure of CO2 at theequilibrium = 0.220 atm

⇒V = the volume of the container = 7.0 L

⇒R = the gas constant = 0.08206 L*atm / mol * K

⇒T = the temperature = 385 K

n = 0.220*7.0/(0.08206*385) = 0.0487 (mol)

this is the amount of CaCO3 which has been converted to CaO before pumping-in additional 0.225 atm CO2(g).

Step 4: Calculate moles CaCO3

After adding additional 0.250 atm CO2(g), the equilibrium CO2 pressure is still 0.220 atm.  All this additional CO2 would completely convert to CaCO3:

n = PV/RT = 0.250*7.0/(0.08206*385) = 0.0554 moles

The total CaCO3 after equilibrium is reestablished is:

0.100 - 0.0487+ 0.0554 = 0.1067 mol

Step 5: Calculate mass CaCO3

Mass CaCO3 = 0.1067 moles * 100.09 g/mol

Mass CaCO3 = 10.68 grams

The final mass of CaCO3 is 10.68 grams

8 0
3 years ago
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