All categories

2 years ago

A student would like to test the effect of eating a lot of candy on

Chemistry

2 answers:

Marina CMI [18]2 years ago

8 0

Answer:

Just to make it short, its eating a lot of candy

Explanation:

Nady [450]2 years ago

4 0

Answer:

Number of candies

Explanation:

The independent variable in this experiment is the number of candies eaten by the student.

An independent variable is best pictured as the cause in an experiment. It is the reason why the effects in experiments are produced.

In this experiment, the hyperactivity is the hyperactivity is the dependent variable. It is the effect that is produced in this set up.

You might be interested in

Which to which! Ahhhh.

jarptica [38.1K]

Answer:

1.a

2. b

Explanation:

Distillation is a process whereby a mixture of liquids having different vapor pressures is separated into its components. At first one might think that this would be quite simple: if you have a solution consisting of liquid A that boils at 50°C and liquid B with a boiling point of 90°C, all that would be necessary would be to heat the mixture to some temperature between these two values; this would boil off all the A (whose vapor could then be condensed back into pure liquid A), leaving pure liquid B in the pot. But that overlooks that fact that these liquids will have substantial vapor pressures at all temperatures, not only at their boiling points.

source: https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chem1_(Lower)/08%3A_Solutions/8.09%3A_Distillation

3 0

11 months ago

List the following elements in order of increasing number of valence electrons: C, CI, As, Na, He.

lorasvet [3.4K]

B is the correct answer

4 0

3 years ago

Find the empirical formula of the following compounds:

Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus $\frac{mol phosphorus}{ 30.97 g phosphorus}$ = 0.0293 mol

Moles bromine 6.99 g bromine $\frac{mol bromine}{79.90 g bromine}$ =0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

8 0

11 months ago

When I2 and FeCl2 are mixed together, iodine (I) cannot replace chlorine (Cl) in the compound because iodine is lower on the per

andrezito [222]

The given statement is True.

This is single replacement reaction.

As reactivity decreases down the group,

Iodine is less reactive than Chlorine.

And so Iodine cannot replace chlorine from FeCl2.

I2 + FeCl2 ---> no reaction (Iodine is less reactive than chlorine)

7 0

2 years ago

Read 2 more answers

What is the new concentration of a solution of CaSO3 if 10.0 mL of a 2.0 M CaSO3 solution is diluted to 100 ml?

kifflom [539]

Answer: The new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

Explanation:

Given: $V_{1}$ = 10.0 mL, $M_{1}$ = 2.0 M

$V_{2}$ = 100 mL, $M_{2}$ = ?

Formula used to calculate the new concentration is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\10.0 mL \times 2.0 M = M_{2} \times 100 mL\\M_{2} = 0.2 M$

Thus, we can conclude that the new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

5 0

2 years ago