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Bingel [31]
3 years ago
7

I NEED HELP ASAP!! What is the difference between the experimental group and a control group?​

Chemistry
1 answer:
dexar [7]3 years ago
3 0

I believe the control group is what doesn't change in the experiment, and the experimental group is what is being tested / receives the treatment :)

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In chemistry what is nickle (II) nitrate solution mixed with a sodium hydroxide solution
AlladinOne [14]

Answer:

I hope this will help you and Please mark me as Brilliant

Explanation:

Approximately 2 mL of Solution A (on the left) is added to a sample of Solution B (on the right) with a dropping pipet. If a precipitate forms, the resulting precipitate is suspended in the mixture. The mixture is then stirred with a glass stirring rod and the precipitate is allowed to settle for about a minute.

Solution A: 0.5 M sodium hydroxide, colorless

Solution B: 0.2 M nickel(II) nitrate, green

Precipitate: light green

Ni(NO3)2(aq) + 2 NaOH(aq) —> Ni(OH)2(s) + 2 NaNO3(aq)

Credits: 

Design

Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859

John W. Moore University of Wisconsin - Madison, Madison, WI 53706

Video

Jerrold J. Jacobsen University of Wisconsin - Madison, Madison, WI 53706

Text

Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859

6 0
3 years ago
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described b
Nostrana [21]

Answer:

0.676 grams of manganese (IV) oxide should be added.

Explanation:

Moles of chlorine gas = n

Volume of the chlorine gas = V = 205 mL = 0.205 L

Pressure of the chlorine gas = 705 Torr = \frac{705}{760}atm=0.928 atm

1 atm = 760 Torr

Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K

PV=nRT ( ideal gs equation)

n=\frac{PV}{RT}=\frac{0.928 atm\times 0.205 L}{0.0821 atm L/mol K\times 298 K}=0.00777 mol

MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)

According to reaction, 1 mole of chlorine gas is obtained from  1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :

\frac{1}{1}\times 0.00777 mol=0.00777 mol of manganese (IV) oxide

Mass of 0.00777 moles of manganese (IV) oxide:

0.00777 mol × 87 g/mol = 0.676 g

0.676 grams of manganese (IV) oxide should be added.

3 0
3 years ago
How many isomers are there in C7H16 ?<br><br>a. 6<br>b. 7<br>c. 8<br>d. 9​
Mariulka [41]
The correct answer is 9 .
7 0
3 years ago
55 kg of liquefied natural gas (lng) are stored in a rigid, sealed 0.17 m3 vessel. in this problem, model lng as 100% methane. d
dexar [7]

I am assuming that the problem ask for the pressure in the system. To be able to calculate this, we first assume that the system acts like an ideal gas, then we can use the ideal gas equation to find for pressure P.

P V = n R T

where,

P = Pressure (unknown)

V = 0.17 m^3

n = moles of lng / methane

R = gas constant = 8.314 Pa m^3 / mol K

T = 200 K

We find for the moles of lng. Molar mass of methane = 16 kg / kmol

n = 55 kg / 16 kg / kmol

n = 3.44 kmol CH4 = 3440 mol

 

Substituting all the values to the ideal gas equation:

P = 3440 mol * (8.314 Pa m^3 / mol K) * 200 K / 0.17 m^3

P = 33,647,247 Pa

<span>P = 33.6 MPa</span>

6 0
3 years ago
Which of these is a drawback of mining
Mamont248 [21]

Answer:

Can you add the choices?

Explanation:

4 0
3 years ago
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