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3 years ago

PLS ANSWER ASAP, WILL GIVE BRAINLIEST ANSWER

Chemistry

1 answer:

Stolb23 [73]3 years ago

6 0

Answer:

.14L or 140mL

Explanation:

This is a classic plug-n-chug problem. Your textbook probably goes over this formula as $M_{1}V_{1} = M_{2}V_{2}$ . M stands for molarity of the given substance, and V stands for the volume that the substance occupies.

Simply plug in the values that you're given, like so:

$M_{1} = 1.5M\\V_{1} = ???\\M_2 = 7M\\V_2 = .03L$

$1.5M * x = 7M * .03L$

After completing the algebra portion and solving for the unknown, you will be left with x = .14L, which is the volume required to neutralize 30mL of 7M NaOH.

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The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

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Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

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where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

Now put all the given values in this formula, we get

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

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