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Crazy boy [7]
1 year ago
12

What is a control group used for?

Chemistry
1 answer:
jeka941 year ago
3 0
  • The control group defined as  a group that does not include any change to the inconstant being tested. Why is a control crucial in an experiment? The control group is crucial because it acts as a benchmark to variatethe results of the experiment to. The experimental group is the group that the scientist is examing . The experimental group receives a change to a variable, or the conditions allowed to change in the experiment.
  • There are two types of control group :   Positive control groups and Negative control groups
  • The positive control group is a group that is contrive to produce the effect you are looking for in the experimental group. The positive control group shows the scientists that the craved results are achievable . This helps stop false negative outcome in the experimental group, where a negative result is receive but is due to a failure in the experiment instead than a truly negative result based on the experimental conditions.
  • A negative control group is a group that is not exposed to the different  scientist is testing, called the independent variable. A negative control group function as a benchmark to secure that the results that are got are actually due to the independent variable and not anything else

To know more about control group visit :

brainly.com/question/14308149

#SPJ9

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When there is an imbalance in a body system, and the body cannot maintain homeostasis, how might other systems respond?
Ulleksa [173]
Other systems in the body might respond with cramps or overheating, which might eventually kill you, if not taken care of in time. Hope this helps!
8 0
3 years ago
Is the perceived frequency of a sound wave
ryzh [129]

Answer:

pitch

Explanation:

7 0
3 years ago
As a real gas deviates from ideal gas behavior, the particles _____
Alik [6]

Answer:

Have some attraction towards each other

Explanation:

Gases deviate from the ideal gas behavior because their molecules have forces of attraction between them. At high pressure, the molecules of gases are very close to each other so the molecular interactions start operating and these molecules do not strike the walls of the container with full impact.

Hope this helps :-)

Have a great rest of your day or night!

Enjoy your studies and assignments

<3 simplysun

ps. I do not own any of these answers so please don't give full credit to me

7 0
3 years ago
1. Using the Slater rule, determine the effective nuclear charge of platinum.
AleksandrR [38]

Answer:

Z* = 3.55

Explanation:

Slater rule says that:

Z*= Z - S

Z* be the nuclear effective charge

Z is the nuclear charge

S is the shielding constant

First we write the electronic configuration of platinum:1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14} 5d^{9} 6s^{1}

The first Slater rule says that we need to group:

(1s^{2}) (2s, 2p)^{8} (3s, 3p)^{8} (3d^{10}) (4s, 4p)^{8} (4d^{10}) (5s, 5p)^{8} (4f^{14}) (5d^{9}) (6s^{1})

The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.

For the third rule we have two considerations, if is ns or np and if is nd or nf:

For our case, we have an electro that is in ns, so the rule says that

-electrons within same group shield 0.35, except the 1s which shield 0.30

-electrons within the n-1 group shield 0.85

-electrons within the n-2 or lower groups shield 1.00

Now we can proceed with the calculation:

The first consideration in the third rule does not apply as we only have one electron on this level.

The second consideration will be as follow for the level 5, where we have 17 electrons.

Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.

So the result for S=(60*1.00 + 17*0.85) = 74.45

And the equation is: Z* = 78 - 74.45

So Z* = 3.55

3 0
3 years ago
Ammonia and oxygen react to form nitrogen and water.
Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0
3 years ago
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