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2 years ago

What is a control group?

Chemistry

2 answers:

kkurt [141]2 years ago

7 0

Answer:

The control group is defined as the group in an experiment or study that does not receive treatment by the researchers and is then used as a benchmark to measure how the other tested subjects do.

Explanation:

The purpose of a control group is to ensure that the independent variable you are testing is really causing the outcomes in the experimental group.

zalisa [80]2 years ago

3 0

Answer: D

Explanation:

A control group is the variable that remains constant throughout the whole experiment

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Which of the following is true of carbon?

pashok25 [27]

I’m pretty sure it’s A or C

8 0

3 years ago

Read 2 more answers

PLEASE HELP! Class is almost ending, I don't want to repeat.

nikklg [1K]

ANSWERS:

1. $2KClO_{3}$ ⇒ 2KCl + $3O_{2}$

2. $H_{2} C_{2} O_{4} + 2NaOH$ ⇒ $Na_{2} C_{2} O_{4} + 2 H_{2} O$

3. $2 C_{10} H_{22} + 31O_{2}$ ⇒ $20C O_{2} + 22 H_{2} O$

4. $4 H_{2} + Fe_{3} O_{4}$ ⇒ $3Fe +4 H_{2} O$

5. $4P + 5 O_{2}$ ⇒ $2 P_{2} O_{5}$

I have attached my work for 1-3. I would like to see if you can get 4 and 5 on your own. but, if you are struggling/copnfused please let me know in the comments! :)

6 0

3 years ago

Imagine that you have a 5.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen

Lorico [155]

Answer:

77.14 atm of pressure should be of an acetylene in the tank.

Explanation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

According to reaction, 2 moles of acetylene reacts with 5 moles of oxygen.

Moles of oxygen= $n_1$

Moles of acetylene = $n_2$

$\frac{n_1}{n_2}=\frac{5 mol}{2 mol}=\frac{5}{2}$

Volume of large tank with oxygen gas, $V_1 = 5.00 L$

Pressure of the oxygen gas inside the tank = $P_1=135 atm$

$RT=\frac{P_1V_1}{n_1}$ ..[1]

Volume of small tank with acetylene gas , $V_2= 3.50 L$

Pressure of the acetylene gas inside the tank = $P_2=?$

$RT=\frac{P_2V_1}{n_2}$ ..[2]

Considering both the gases having same temperature T, [1]=[2]

$\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}$

$P_2=\frac{P_1V_1\times n_2}{V_2\times n_1}$

$=\frac{135 atm\times 5.00 L\times 2}{3.50 L\times 5}=77.14 atm$

77.14 atm of pressure should be of an acetylene in the tank.

8 0

3 years ago

Match each of the substances with the expected pH of the resulting solution when they are added to neutral water. Determine if a

Tamiku [17]

To determine the expected pH of the resulting solution of the following substances, create a balanced chemical equation of their ionization in water:

HI

HI + H2O ---> H+ + I-

It completely dissociates into H+ and I-. Due to the presence of the Hydronium Ion, the solution is acidic.

KBr

KBr + H2O ----> HBr + KOH

The salt KBr is formed by a strong base and a weak acid, therefore, the solution it forms with water is basic.

LiOH

LiOH + H2O ----> Li+ + OH-

It dissociates completely in water, turns into Li+ and OH-. Due to the presence of Hydroxide Ion, the solution becomes basic.

3 0

2 years ago

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

vovikov84 [41]

Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .

Hence, the empirical formula of this compound would be $\rm C_2H_8N_2O_3$ .

6 0

3 years ago