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zloy xaker [14]
3 years ago
6

Which components can be found in both comparative and experimental investigations, but are never found in descriptive investigat

ions? Check all that apply.
- question
- hypothesis
- variables
- control group
- prediction
- conclusion
Chemistry
2 answers:
Reptile [31]3 years ago
6 0

Answer:

HYPOTHESIS, VARIABLES, AND CONTROL GROUP

Explanation:

YOUR WELCOME :)

alex41 [277]3 years ago
4 0

Answer:

Hypothesis and Variables

Explanation:

I just took the quiz

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Any mixture that is heterogeneous on a microscopic level is a
Nadya [2.5K]
Solution is the answer.
5 0
4 years ago
What is the name of this compound? H single bonded to N with a pair of electron dots above and a single bond to H below, single
dybincka [34]

Answer:

Ethanamine (also known as ethylamine)

Explanation:

The compound that is requested by the question is ethanamine. Its trivial name is ethylamine.

It is a compound that contained the ethyl moiety (CH3CH2-) as well as the amine moiety (-NH2).

Ethanamine has a structure that can easily be determined by the statements in the question.

The structure of ethanamine is shown in the image attached.

4 0
3 years ago
B. What useful functions do oxidation numbers
disa [49]

Explanation:

b. What useful functions do oxidation numbers  serve?

It is used to show oxidation and reduction (loss and gain of electrons)

b. How many molecules are in 1 mole of  molecules?

1 mole = 6.022 * 10^23 molecules

c. What is the name given to the number of  molecules in 1 mole?

Avogadro's Number of molecules

21. a. What is the molar mass of an element?

This is the mass of an element divided by the number of moles.

Molar mass = Mass / Number of moles

b. Write the molar mass rounded to two  decimal places of carbon, neon, iron and  uranium.

amu = Atomic Mass Unit

Carbon = 12.01 amu

Neon = 20.18 amu

Iron = 55.85 amu

Uranium = 238.03 amu

7 0
3 years ago
URGENT !! A substance has 55.80% carbon, 7.04% Hydrogen, and 37.16% Oxygen. What is it's empirical and molecular formula if it h
Ede4ka [16]
<h3><u>Answer;</u></h3>

Empirical formula = C₂H₃O

Molecular formula = C₁₄H₂₁O₇

<h3><u>Explanation</u>;</h3>

Empirical formula

Moles of;

Carbon = 55.8 /12 = 4.65 moles

Hydrogen = 7.04/ 1 = 7.04 moles

Oxygen  = 37.16/ 16 = 2.3225 moles

We then get the mole ratio;

4.65/2.3225 = 2.0

7.04/2.3225 = 3.0

2.3225/2.3225 = 1.0

Therefore;

The empirical formula = <u>C₂H₃O</u>

Molecular formula;

(C2H3O)n = 301.35 g

(12 ×2 + 3× 1 + 16×1)n = 301.35

43n = 301.35

  n = 7

Therefore;

Molecular formula = (C2H3O)7

                             <u> = C₁₄H₂₁O₇</u>

6 0
4 years ago
Please help me with this
Harrizon [31]
I and Ca is the answer
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