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Nitella [24]
3 years ago
14

What is the experimental group? on edpuzzle

Chemistry
2 answers:
RideAnS [48]3 years ago
3 0
An experimental group, also known as a treatment group, receives the treatment whose effect researchers wish to study, whereas a control group does not. They should be identical in all other ways
elixir [45]3 years ago
3 0

Answer: An experimental group is a test sample or the group that receives an experimental procedure.

Explanation: This group is exposed to changes in the independent variable being tested. The values of the independent variable and the impact on the dependent variable are recorded. An experiment may include multiple experimental groups at one time.

Hope this helps! :)

You might be interested in
Which of these can be combined to make a mixture?
Brrunno [24]
Two liquids is the correct answer
5 0
2 years ago
Read 2 more answers
If 1.20 g of Zn is allowed to react with 2.00 g of CuSO4, according to the equation below, how many grams of Zn will remain afte
sdas [7]

Answer:

After the reaction is complete 0.94 g of Zn are still remaining.

Explanation:

We define the reaction as:

CuSO₄ (aq)  +  Zn (s)  →  ZnSO₄ (aq)  +  Cu(s)  

In order to determine the grams of Zn that will remain after the reaction is complete, we need to confirm if the Zn is the limiting reactant.

We determine moles of each reactant:

1.20 g / 65.41 g/mol = 0.0183 moles of Zn

2 g / 159.6 g/mol = 0.0125 moles of sulfate

Ratio is 1:1. 1 mol of sulfate needs 1 mol of Zn to react.

If I have 0.0125 moles of sulfate I will ned the same moles of Zn to complete the reaction.

I have 0.0183 moles, so it is ok that Zn is the reagent in excess

We convert the moles to mass and we make the substraction to answer the grams of Zn that will remain after the reaction:

0.0183 moles . 159.1 g /mol = 2.92 g

0.0125 moles . 159.1 g /mol =  1.98 g

To complete all the reaction we need 2.92 g of sulfate, but I have 1.98 g

After the reaction is complete (2.92 g - 1.98 g) = 0.94 g are still remaining.

6 0
3 years ago
1. Name the mode of heat transfer
s344n2d4d5 [400]

Answer:

there are three process of heat transfer:conduction,convection,radiation

6 0
3 years ago
Describe the chemical reaction based on the chemical equation below. Also, explain whether the equation is balanced.
liubo4ka [24]

Answer:

The balanced equation is: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(I)

4 moles of ammonia need 5 moles of oxygen to react, in order to produce 6 moles of water and 4 moles of nitrogen oxide.

Explanation:

NH₃(g) + O₂(g) → NO(g) + H₂O(I)

That is the unbalanced equation.

Ratio is 1:1, so 1 mol of ammonia and 1 mol of oxygen may react to produce 1 mol of NO and 1 mol of water

First of all we need to balance. Let's count the atoms:

In reagent side, we have 3 H and in product side there are 2H, we can add a 3, so now we have 6 H. Therefore, the reagent side must have 2 in front of the ammonia. In both sides we have 6 H

Now, that we added a 2 in the ammonia, we have 2N, so we have to add 2 to NO in the product side.

Let's count the oxygen in the product side, 5. We must add 5/2 in the reactant side. But chemical equations usually do not have rational numbers, that's why we multiply x2, all the stoichiometry.

The balanced equation is: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(I)

4 moles of ammonia need 5 moles of oxygen to react, in order to produce 6 moles of water and 4 moles of nitrogen oxide.

3 0
3 years ago
Read 2 more answers
Calculate the pH in an aqueous 0.120 M nitrous acid solution.
Anvisha [2.4K]

Answer:-  pH is 2.14.

Solution:- Nitrous acid, HNO_2 is a weak acid so first of all we solve for H_3O^+ and then figure out the pH.

the equation is written as:

HNO_2(aq)+H_2O(l)\leftrightharpoons H_3O^+(aq)+NO_2^-(aq)

Initial concentration for the acid is given as 0.120 M. Let's say the change in concentration is x. Then the equilibrium concentrations would be as:

HNO_2=0.120-x

H_3O^+ = x

NO_2^- = x

Ka for nitrous acid is 4.5*10^-^4 and the equilibrium expression for this would be written as:

Ka=\frac{[H_3O^+][NO_2^-]}{HNO_2}

Let's plug in the values in it.

4.5*10^-^4=\frac{(x)^2}{0.120-x}

To make the calculations easy we could ignore x for the bottom and the expression becomes:

4.5*10^-^4=\frac{(x)^2}{0.120}

On cross multiply:

x^2=4.5*10^-^4*0.120

On taking square root to both sides:

x=7.3*10^-^3

So, [H_3O^+]=7.3*10^-^3M

Now we could calculate the pH using the pH formula:

pH=-log[H_3O^+]

pH=-log(7.3*10^-^3)

pH = 2.14

So, the pH of 0.120M nitrous acid is 2.14.

8 0
2 years ago
Read 2 more answers
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