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What is the connection between Archimedes and the Mass-Volume lab

Usimov [2.4K]

Answer:

Explanation:

This question is not very clear, however, going by the key words in the question, Archimedes' principle shows a relationship between mass and volume.

Archemedes' principle states that when a body is fully or partially immersed in fluid, it is acted upon by an upward force which is equal to the weight of the fluid displaced by the body.

The volume displaced here can be determined using the formula

Density of liquid = mass displaced ÷ volume

Hence, volume = mass displaced ÷ density of liquid

4 0

3 years ago

Which of these solutions would be considered to have a neutral pH?

Citrus2011 [14]

Answer:

Pure Water.

Explanation:

Solutions with a pH of 1-6 are considered acidic. Solutions with a pH of 8-14 are considered basic. Solutions with a pH of 7 are considered neutral.

6 0

3 years ago

What does the symbol ka mean in chemistry?

Dmitry_Shevchenko [17]

<span>Ka is an equilibrium constant for the partial ionization of "weak" acids in water.</span>

4 0

4 years ago

Read 2 more answers

can you help me I'm trying to do a lm trying to do a keynote and i need to know what it means and what it dose

yanalaym [24]

Here are my quick example slides:

1st Picture:
just a title page

2nd Picture:
I just gave the definition of an organelle you could go more in-depth. The picture doesn't have to be there, of course.

3rd Picture:
Examples of organelles in plant and animal cells, these are just examples of how I would do one.

This was just to show how you could do it because it would be kind of hard for me to explain just in words.

Hope I could help! c:

8 0

3 years ago

If 24.3 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over?2 NO (g) + O₂ (g) → 2

zhuklara [117]

Explanation:

2 NO (g) + O₂ (g) ----> 2 NO₂ (g)

24.3 g of NO are reacting with 13.8 g of O₂. First we can convert the mass of theses samples into moles using their molar masses.

molar mass of O = 16.00 g/mol

molar mass of N = 14.01 g/mol

molar mass of NO = 16.00 g/mol + 14.01 g/mol

molar mass of NO = 30.01 g/mol

molar mass of O₂ = 2 * 16.00 g/mol

molar mass of O₂ = 32.00 g/mol

moles of NO = 24.3 g * 1 mol/(30.01 g)

moles of NO = 0.810 moles

moles of O₂ = 13.8 g * 1 mol/(32.00 g)

moles of O₂ = 0.431 moles

Now, to determine the limiting reactant or the excess reactant we can find the number of moles of O₂ that will react with 0.810 moles of NO and the number of moles of NO that will react with 0.431 moles of O₂.

According to the coefficients of the reaction 2 moles of NO will react with 1 mol of O₂. Let's use that relationship to find the limiting reagent.

2 moles of NO = 1 mol of O₂

moles of O₂ = 0.810 moles of NO * 1 mol of O₂/(2 moles of NO)

moles of O₂ = 0.405 moles

moles of NO = 0.431 moles of O₂ * 2 moles of NO/(1 mol of O₂)

moles of NO = 0.862 moles

We found that we need 0.405 moles of O₂ to completely react with 0.810 moles of NO. Or, we need 0.862 moles of NO to completely react with ours 0.431 moles of NO.

We can say that NO is limiting our reaction and O₂ is in excess.

Only 0.405 moles of O₂ will react with 0.810 moles of NO. But we had 0.431 moles of it. Let's find the excess.

Excess of O₂ = 0.431 moles - 0.405 moles

Excess of O₂ = 0.026 moles

Answer: 0.026 moles is the number of moles of oxygen that will be left over.

4 0

1 year ago