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Deffense [45]
4 years ago
5

Which answer below is not an indication that a chemical reaction has occurred?

Chemistry
1 answer:
ankoles [38]4 years ago
4 0
Only when there is no product that is when a chemical reaction has not occured
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Which of the following atoms would have the smallest atomic radius?
e-lub [12.9K]

Answer:

Lithuim is the smallest

Explanation:

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3 years ago
A golf ball and bowling ball are moving and both have the same kinetic energy.Which one is moving faster? If they move at the sa
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I believe the answer to the first question is the golf ball because it requires less energy to move it so it will move faster than the bowling ball will with less energy. For the second question I could be wrong but I think the bowling ball has more kinetic energy because according to the equation KE= 0.5*m* x^{2} where KE represents the kinetic energy the m is the mass and x is the speed of the object the bowling ball will have more energy.
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3 years ago
Find how many milliliters of NaOH should be used to reach the half-equivalence point during the titration of 20.00 mL 0.888 M bu
Varvara68 [4.7K]

<u>Answer:</u> The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

<u>Explanation:</u>

The chemical equation for the dissociation of butanoic acid follows:

CH_3CH_2CH_2COOH\rightleftharpoons CH_3CH_2CH_2COO^-+H^+

The expression of K_a for above equation follows:

K_a=\frac{[CH_3CH_2CH_2COO^-][H^+]}{[CH_3CH_2CH_2COOH]}

We are given:

[CH_3CH_2CH_2COOH]=0.888M\\K_a=1.54\times 10^{-5}

[CH_3CH_2CH_2COO^-]=[H^+]

Putting values in above expression, we get:

1.54\times 10^{-5}=\frac{[H^+]^2}{0.888}

[H^+]=-0.0037,0.0037

Neglecting the negative value because concentration cannot be negative

To calculate the volume of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is butanoic acid

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=\frac{0.0037}{2}M\text{ (half equivalence)}\\V_1=20.00mL\\n_2=1\\M_2=0.425M\\V_2=?mL

Putting values in above equation, we get:

1\times \frac{0.0037}{2}\times 20.00=1\times 0.425\times V_2\\\\V_2=\frac{1\times 0.0037\times 20}{1\times 0.425\times 2}=0.087mL

Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

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