All categories

1 year ago

What are some possible bias in an experiment about how sleep affects mood HELP ASAPPPPP

1 answer:

7 0

Answer: Some possible bias in an experiment about how sleep affects mood could be a person's average mood with or without sleep, how people are acting in the experiment compared to how they actually would in their normal environment, or even how the person collected the data for the experiment interprets their moods.

Explanation:

1. A person's mood could be relatively different depending on other factors than just sleep, so it could affect experiment. For example if an emotional crisis had occurred recently, it could affect your mood and environmental factors could affected your mood, like weather.

2. A person could also change their mood around strangers or even when participating in the experiment. For instance if they aren't comfortable with the person taking down the analysist or the other test subjects, they make act differently than they usually do at home where they're comfortable.

3. A person collecting the data will have to know how to specify and determine a persons mood with their instruments. For example, if they're doing it just by talking with the test subjects, they'll have to know how to properly understand their different stages of each of their moods with out the data collectors mood changing as well. Also, if the data collectors mood is changed at all it can slightly affect the experiments results as well.

You might be interested in

A piece of sodium metal can be described as?

vlada-n [284]

Answer:

A piece of sodium metal can be described as a pure substance and an element. ... Sodium is a soft, waxy, silvery, highly reactive metal. Sodium belongs to group 1 of the periodic table and it is highly abundant in the earth's crust. Sodium can also be found in various minerals such as rock salt, feldspars and sodalite.

Explanation:

here is short description that will help u 2 understand more

Thx

6 0

2 years ago

Read 2 more answers

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

2 years ago

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Svet_ta [14]

Answer:

$V_1=23.3~mL$

Explanation:

In this case, we have a dilution problem. We have to remember that in the dilution procedure we go from a solution with higher concentration to a solution with lesser concentration. Therefore we have to start with the dilution equation:

$C_1*V_1=C_2*V_2$