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Eva8 [605]
2 years ago
12

Which term is the symbol for the rate constant? A. [A] B. k C. R D. x + y E. None of the Above

Chemistry
1 answer:
Artemon [7]2 years ago
3 0
Answer:
            Option-B (k) is the correct answer.

Explanation:
                   As we know the rate of reaction is given as;

                                                Rate  =  k [A]ˣ
Where;
            Rate  =  Rate of Reaction

            k  =  rate constant
 
            [ ]  =  concentration of A

            x  =  order of reaction

So, from this equation we found that rate of reaction depends upon concentration and rate constant (k).
Now,
The rate constant is as follow,,

                                            k  =  Ae^(Ea/RT) 

This equation is known as Arrhenius Equation, according to this equation rate constant depends upon Temperature and Activation energy. Greater the temperature greater is the rate constant and hence greater is the rate of reaction. Or smaller the activation energy greater is the rate constant and vice versa.
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You collect 552 mL of argon gas at 23.0 C. What volume will the gas occupy at 46.0C if the pressure remains constant?
taurus [48]

Answer:

1027.9 mL

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

Explanation:

Formula P1 x V1 / T1 = P2 x V2 / T2

Fill in what you know

Pressure is constant so no need to put that in making the formula

V1 / T1 = V2 / T2

Voulme 1= 950 mL

Volume 2= ?

Temperature 1 = 25 C

Temperature 2 = 50 C

8 0
2 years ago
For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _
nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.  

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g).

When \rm H_2 is added to \rm H_2C\text{=}CH_2 (ethene) under heat and with the presence of a catalyst, \rm H_3C\text{-}CH3 (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, \Delta S < 0.

The equation for the change in Gibbs Free Energy for a particular reaction is:

\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}}).

For a particular reaction, the more negative \Delta G is, the more spontaneous ("favorable") the reaction would be.

Since typically \Delta S < 0 for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

T (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

8 0
3 years ago
Calculate the molarity of 139 grams of sucrose, C12H22O11, in 2.60 liters of solution.
Allisa [31]

Answer:

to calculate the molarity of the said sucrose,

firstly calculate the moles

which is = Molecular weight of C12H22O11 = 342g/mol

then

moles = 139/342

= 0.41 moles

to calculate Molarity now

Molarity= moles of the solute/volume of solution in liter

=0.41/2.60

=0.158M

Explanation:

4 0
3 years ago
How many moles of LiF would be required to produce a 2.5 M solution with a volume of 1.5L
mihalych1998 [28]

Multiply the values.

n = cv

Where

n =mol

c = concentration

v = volume

n = 2.5×1.5 = 3.75mol

3 0
3 years ago
The constitutional isomer of ethanol, dimethyl ether (CH3OCH3), is a gas at room temperature. Suggest an explanation for this ob
Sedaia [141]

Answer:

Because of its weak intermolecular forces.

Explanation:

Hello there!

In this case, according to the given description, it turns out possible for us to recall the chemical structures of both ethanol and dimethyl ether as follows:

CH_3CH_2OH\\\\CH_3COCH_3

Thus, we can see that ethanol have London dispersion forces (C-C bonds), dipole-dipole forces (C-O bonds) and also hydrogen bonds (O-H bonds) which make ethanol a liquid due to the strong hydrogen bonds. On the other hand, we can see that dimethyl ether has just London and dipole forces, which are by far weaker than hydrogen bonding, that makes it unstable when liquid and therefore it tends to vaporize quite readily.

Regards!

8 0
3 years ago
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