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4 years ago

Write an investigative question

Chemistry

1 answer:

pshichka [43]4 years ago

6 0

Answer:

What happened?

What was the date, time and duration of the incident or behavior?

How many times did this happen?

Where did it happen?

How did it happen?

Did anyone else see it happen? Who? What did they say? What did they do?

Was there physical contact? Describe it. Demonstrate it.

What did you do in response to the incident or behavior?

What did you say in response to the incident or behavior?

How did the subject of the allegation react to your response?

Did you report this to anyone in management? To whom? When? What they say and/or do?

Did you tell anyone about the incident or behavior? Who? What did they say and/or do?

Do you know whether the subject of the allegation has been involved in any other incidents?

Do you know why the incident or behavior occurred?

Do you know anyone else who can shed light on this incident?

Is there anything else you want to tell me that I haven’t asked you?

Explanation:

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6) A student measures out 96.21 g of sulfur for an experiment. How many moles of Sulfur are in this

barxatty [35]

Atomic mass of Sulfur = 32g

32g of Sulfur is one mole.

1g of Sulfur is $\frac{1}{32} moles$

96.21g of Sulfur is $\frac{96.21}{32} moles=> 3moles(appx)$

5 0

3 years ago

Read 2 more answers

What is respiration? Explain in your own words

Black_prince [1.1K]

Answer:

breathing

Explanation:

Your welcome lol

hope this helps :) pls give brainliest

4 0

2 years ago

Read 2 more answers

A chemistry student is given 600. mL of a clear aqueous solution at 37.° C. He is told an unknown amount of a certain compound X

barxatty [35]

Answer:

Yes, it is 14. g of compound X in 100 ml of solution.

Explanation:

The relevant fact here is:

the whole amount of solute disolved at 21°C is the same amount of precipitate after washing and drying the remaining liquid solution: the amount of solute before cooling the solution to 21°C is not needed, since it is soluble at 37°C but not soluble at 21°C.

That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.

Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.

With that, the solubility can be calculated from the followiing proportion:

84. g solute / 600 ml solution = y / 100 ml solution

⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.

The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.

The answer is 14. g of solute per 100 ml of solution.

7 0

3 years ago

this element has an atomic number lower than that of aluminum and one less valence electrons then the Group 16 elements

ahrayia [7]

This element is nitrogen (N)

6 0

3 years ago

The reaction: 2 CO(l) + O2(g) ⇄2 CO2(g) with a H = −25.0 kJ/mol, is at equilibrium. Which of the following lists three ways thi

elena-s [515]

Answer:

Option B and D both have 3 ways to shift the reaction to the right

Explanation:

The reaction 2 CO(l) + O2(g) ⇄2 CO2(g) is exothermic because ΔH = -25 kJ < 0 This means heat will be released

Therefore, increasing the temperature will shift the equilibrium to the left, while decreasing the temperature will shift the equilibrium to the right.

A. remove CO2, lower pressure and remove CO

⇒ Lowering the pressure will shift the equilibrium to the side with most moles of gas : On the left side there are 3 moles, on the right side 2 moles. By lowering the pressure, the equilibrium will shift to the left.

B remove CO2, raise pressure and add CO

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the right side.

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The right side).

⇒ Add CO: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added CO. (The right side).

C raise temperature, lower pressure and remove O2

⇒ Increasing the temperature will shift the equilibrium to the left

D add O2, raise pressure and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the right

⇒ By raising the pressure, the equilibrium will shift to the side with the lesser amount of moles of gas. This is the right side.

⇒ Add O2: the equilibrium will shift to the side of CO2, so the reaction will shift toward the products side to reduce the added O2. (The right side).

E remove CO2, increase volume and lower temperature

⇒ decreasing the temperature will shift the equilibrium to the right

⇒ Remove CO2: the equilibrium will shift to the side of CO2 ,so the reaction will shift toward products to replace the product removed. (The right side).

⇒ Increase volume : with a pressure decrease due to an increase in volume, the side with more moles is more favorable. The equilibrium will shift to the left side.

8 0

3 years ago