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cluponka [151]
3 years ago
6

Thinking and questioning is the start of the scientific inquiry process. Please select the best answer from the choices provided

T F
Chemistry
2 answers:
aev [14]3 years ago
8 0
Yes thats true! You always have to think about the question or project before you start a science experiment! :) 




SashulF [63]3 years ago
6 0

Answer: TRUE

Explanation:

A scientific inquiry is a coordinated and thoughtful process to search, explain and predict the cause of natural phenomena. It is associated with steps such as framing a question, data collection, analysis and interpretation. In this way a research design is required to be created on the basis of the thinking and creativity of the scientists.

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A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
When chlorobenzene reacts with Mg in ether followed by CO2 and neutralization with dilute HCl, __________ will be formed.
allsm [11]

Answer:

c. benzoic acid

Explanation:

The given reaction is an example of a Grignard reaction:

When chlorobenzene (C₆H₅Cl) reacts with Mg in ether, an intermediate is formed (C₆H₅MgCl).

Said intermediate then reacts with CO₂ producing a benzoic acid salt (C₆H₅CO₂X), this salt is then neutralized with dilute HCl producing benzoic acid (C₆H₅CO₂H).

4 0
3 years ago
Need help asap<br> will give brainly
zheka24 [161]
I think the answers are A, C, A.
7 0
2 years ago
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What occurs during one half-life?
lorasvet [3.4K]
The half-life of a radioactive compound is the time taken for that said isotope to decay or disintegrate so that only half of the initial atoms remain in that compound. During the decay process, the isotope will give off energy and matter, and the way to depict this is indicated by t 1/2.
8 0
3 years ago
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J. J. Thomson’s experiment disproved the theory that an atom
gizmo_the_mogwai [7]

Answer : J. J. Thomson’s experiment disproved the theory that an atom is indivisible.

Explanation : Scientist J.J. Thomson did his experiment to prove the existence of electrons. He did the experiment using a cathode ray tube, in which a vacuum-sealed tube with a cathode and anode on one end was placed which created a beam of electrons that traveled towards the other end of the tube. This was the theory that proved that atoms consists of many subatomic particles namely electrons.

7 0
3 years ago
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