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Studentka2010 [4]
3 years ago
5

Which of the following is an example of observational scientific investigation?

Chemistry
2 answers:
Softa [21]3 years ago
8 0
B: nothing the color of the solution when heated
hoa [83]3 years ago
5 0

Answer:

The correct answer will be option-B.

Explanation:

A scientific investigation involves many steps to answer and explain the natural phenomenon.  The observational scientific investigation is a way which includes making an observation which is also the first step of the scientific method.

In the given question, observing the color of the solution while heating will be considered as an example of observational investigation as it states that something has been observed with the senses of the human. All other are cases of experimental investigation.

Thus, Option-B is the correct answer.

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Which group of elements has the most similar chemical properties? (1 point)
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Answer:

carbon nitrogen and oxygen

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Describe the process used to measure out a specific mass of a solid
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Place the solid on an electronic balance. Read the balance reading.
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A sample of fluorine gas exerts a pressure of 900 mmHg. When the pressure is changed to 1140 mmHg, the volume is 250 mL. What wa
UNO [17]
Let's use Boyle's Law here. P1*V1 = P2*V2
Given: (assuming that there are decimals at the end for Sig Figs)
P1 = 900.mmHg
P2 = 1140.mmHg
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2 years ago
Calcola la massa (in g) di un oggetto d'oro che ha
AfilCa [17]

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3 0
2 years ago
The isotope Np-238 has a half life of 2.0 days if 96 grams of it were present on Monday how much will remain six days later
Kipish [7]

Answer:

12.02 g

Explanation:

From the question given above, the following data were obtained:

Half life (t½) = 2 days

Original amount (N₀) = 96 g

Time (t) = 6 days

Amount remaining (N) =..?

Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:

Half life (t½) = 2 days

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 2

K = 0.3465 /day

Therefore, the rate of disintegration of the isotope is 0.3465 /day.

Finally, we shall determine the amount of the isotope remaining after 6 days as follow:

Original amount (N₀) = 96 g

Time (t) = 6 days

Decay constant (K) = 0.3465 /day.

Amount remaining (N) =.?

Log (N₀/N) = kt / 2.303

Log (96/N) = (0.3465 × 6) / 2.303

Log (96/N) = 2.079/2.303

Log (96/N) = 0.9027

Take the anti log of 0.9027

96/N = anti log (0.9027)

96/N = 7.99

Cross multiply

96 = N × 7.99

Divide both side by 7.99

N = 96 /7.99

N = 12.02 g

Therefore, the amount of the isotope remaining after 6 days is 12.02 g

3 0
2 years ago
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