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AlexFokin [52]
3 years ago
5

During the observation / question phase of the scientific method, a scientist is usually: collecting data creating graphs and/or

performing calculations reporting the results to a colleague none of the above
Chemistry
2 answers:
abruzzese [7]3 years ago
6 0

Answer:

Collecting data  

Step-by-step explanation:

During the observation phase of the scientific method, scientists are collecting data.

They can't create graphs or perform calculations, because they are still collecting data, nor can they report results to a colleague, because they have none to report.

NeTakaya3 years ago
5 0

Answer:

collecting data

Explanation:

Collecting data is a systematic endeavor. There is dependent variable and we are gathering information about that variable. This step helps us to understand how the dependent variable behaves under controlled experimental conditions. Collecting data eventually helps us to test hypothesis, understand research questions and evaluate outcomes.

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What is ancient japan remembered for
Katarina [22]
Hi!

The Samurai is what I would attribute to the memory of Ancient<span> Japan.</span>
3 0
3 years ago
Devise a way to separate a mixture of sand, salt, and iron filings. Write the procedure using a step-by-step method and explain
aev [14]
1. Iron fillings are magnetic, so use a magnet to pull the iron fillings out of the mix.
2. Then you can put the salt and sand mixture into water, since salt is soluble, and the salt will dissolve, leaving you with sand.
3 0
3 years ago
Caed for this question.
OLga [1]

Answer:

0.962 atm.

97.4 kPa.

731 torr.

14.1 psi.

97,434.6 Pa.

Explanation:

Hello.

In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:

- Atmospheres: 1 atm = 760 mmHg:

p=731mmHg*\frac{1atm}{760mmHg} =0.962atm

- Kilopascals:: 101.3 kPa = 760 mmHg:

p=731mmHg*\frac{101.3kPa}{760mmHg} =97.4kPa

- Torrs: 760 torr = 760 mmHg:

p=731mmHg*\frac{760 torr}{760mmHg} =731 torr

- Pounds per square inch: 14.69 psi = 760 mmHg:

p=731mmHg*\frac{14.69}{760mmHg} =14.1psi

- Pascals: 101300 Pa = 760 mmHg:

p=731mmHg*\frac{101300Pa}{760mmHg} \\\\p=97,434.6Pa

Best regards.

5 0
3 years ago
It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta
Zigmanuir [339]

Answer:

\lambda=241.9\ nm

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that  

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,  

1 mole = 6.023\times 10^{23}\ electrons

So,  

6.023\times 10^{23} electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of \frac {495.0}{6.023\times 10^{23}}\ kJ of energy.

Energy required = 82.18\times 10^{-23}\ kJ

Also,  

1 kJ = 1000 J

So,  

Energy required = 82.18\times 10^{-20}\ J

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}

\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}

\lambda=\frac{19.878}{10^6\times \:82.18}

\lambda=2.4188\times 10^{-7}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=241.9\ nm

6 0
3 years ago
If irons temperature goes from 50c to 75c in a 20g sample, how many joules of heat were used?
Marina CMI [18]

Answer:

Option D. 230 J

Explanation:

We'll begin by calculating the temperature change of the iron. This can be obtained as follow:

Initial temperature (T₁) = 50 °C

Final temperature (T₂) = 75 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 75 – 50

ΔT = 25 °C

Thus, the temperature change of the iron is 25 °C.

Finally, we shall determine the amount of heat energy used. This can be obtained as follow:

Mass (M) = 20 g

Change in temperature (ΔT) = 25 °C

Specific heat capacity (C) = 0.46 J/gºC

Heat (Q) =?

Q = MCΔT

Q = 20 × 0.46 × 25

Q = 230 J

Thus, the amount of heat used was 230 J

3 0
3 years ago
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