As the volume of a sample increases from 20. ml to 30. ml, does the mass increase or decrease?

3 answers
Question:

As the volume of a sample increases from 20. ml to 30. ml, does the mass increase or decrease?

Answers

The concentration of the reactants decreases.

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

k= rate constant

x = order with respect to

y = order with respect to

Thus as the concentration of reactants decrease, the rate of the reaction also decrease and thus the rate of formation of products decrease.

2. Which 0.1 M solution contains an electrolyte?

(1) C6H12O6(aq) (3) CH3OH(aq)

(2) CH3COOH(aq) (4) CH3OCH3(aq)

Explanation:

a) P(x>95)=0.32

b) P(M20>95)=0.02

c) P(M30>95)=0.005

d) Increasing the sample size narrows the sampling distribution. Larger samples will have a mean which has more probability of being closer to the population mean. This is reflected by the decrease in the standard deviation of the sample mean when the sample size increases.

e) If a random sample of 30 time intervals between eruptions has a mean longer than 95 minutes, i can conclude that the mean time between eruptions may not be 85 minutes. This sample result is such an unlikely result that it can serve as evidence to question the validity of the mean time between eruptions.

f) Treating these 22 eruptions as a random sample, the likelihood the mean length of time between eruptions exceeds 88.8 minutes is 0.20.

Step-by-step explanation:

We have the the time between eruptions as normally distributed with mean = 85 min. and standard deviation = 21.25.

a) We have to calculate P(x>95)

[tex]z=(X-\mu)/\sigma=(95-85)/21.25=10/21.25=0.47\\\\P(X95)=P(z0.47)=0.32[/tex]

b) The sampling distribution of samples of size n=20 is

[tex]\mu_s=\mu=85\\\\\sigma_s=\sigma/\sqrt{n}=21.25/\sqrt{20}=4.75[/tex]

The probability that the sample mean is longer than 95 is

[tex]z=(X-\mu)/\sigma=(95-85)/4.75=10/4.75=2.11\\\\P(M_{20}95)=P(z2.11)=0.02[/tex]

c) Now, with n=30, the mean and standard deviation becomes:

[tex]\mu_s=\mu=85\\\\\sigma_s=\sigma/\sqrt{n}=21.25/\sqrt{30}=3.88[/tex]

[tex]z=(X-\mu)/\sigma=(95-85)/3.88=10/3.88=2.58\\\\P(M_{30}95)=P(z2.58)=0.005[/tex]

f) The sample size is n=22

[tex]\mu_s=\mu=85\\\\\sigma_s=\sigma/\sqrt{n}=21.25/\sqrt{22}=4.53[/tex]

We have to calculate x so that P(X>x)=0.2

This probability happens for a z=0.84.

Then, we can calculate x as:

[tex]z=\mu+z*\sigma=85+0.84*4.53=85+3.8=88.8[/tex]

Treating these 22 eruptions as a random sample, the likelihood the mean length of time between eruptions exceeds 88.8 minutes is 0.20.

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