The amplitude of a transversal wave in a spring toy triples. What happens to the energy of the wave in thetoy?

Answers

Answer 1

Answer:

Answer: The energy of the wave increases by a factor of 9

Explanation:

3 x 3 = 9

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What 2 pieces of equipment will you use to separate the salt from the water?

Answers

Separate Salt and Water Using Distillation

If you want to collect the water, you can use distillation. This works because salt has a much higher boiling point than water. One way to separate salt and water at home is to boil the salt water in a pot with a lid.

hope this helps!!

Final answer:

To separate salt from water, you can use evaporation or filtration.

Explanation:

To separate the salt from the water, you can use the processes of evaporation and filtration.

Evaporation: You can heat the saltwater solution to boil off the water and leave behind the salt as solid crystals. The water evaporates and can be collected as a vapor if desired.
Filtration: Alternatively, you can use a filter to separate the solid salt particles from the liquid water. The saltwater solution is poured through the filter, which traps the salt while allowing the water to pass through.

Learn more about salt separation here:

brainly.com/question/12827858

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CH₃OH(l) + CO(g) → CH₃CO₂H(aq) What mass in grams of CO would be required for the complete carbonylation of 14.35 grams of methanol to acetic acid?

Answers

CO has a molar mass of 44.01 g/mol, which means that 1 mole of CO weighs 44.01 grams.

We want to convert 14.35 grams of ethanol to acetic acid, so we need to know how many moles of ethanol there are. Let's first multiply the given mass of ethanol by its molar mass:

14.35 (g) x 32.06 (g/mol) = 45.85 mol

So, we have 45.85 moles of methanol. If methanol reacts with CO to produce acetic acid, we need to know how many moles of CO we need to use, based on the stoichiometry of the reaction. The reaction equation gives us a 2:1 ratio for CO and methanol:

2 CH₃OH(l) + CO(g) → CH₃CO₂H(aq)

So, for every mole of CH₃OH we react, we will need 0.5 mole of CO. Therefore, we need 23.92 moles of CO to convert all the methanol to acetic acid.

Using the mass of 1 mole of CO, we can convert moles of CO to grams:

23.92 (mol) x 44.01 (g/mol) = 1050.93 (g)

So, we need 1051 grams of CO to completely convert 14.35 grams of methanol to acetic acid.

The latent heat of vaporization of ethanol is 200 calories/gram. This means that _____. about 200 calories of heat must be added to 1 gram of ethanol to change its temperature about 200 calories of heat must be added to 1 gram of ethanol to convert it from a solid to a liquid about 200 calories of heat must be added to 1 gram of ethanol to convert it from a liquid to a gas

Answers

Heat of vaporization refers to the amount of energy required to vaporize some amount of a substance depending on the units. Since the units in this case is calories/gram, 200 calories are required to vaporize 1 gram of ethanol. This means that it takes 200 calories to convert 1 gram of ethanol from a liquid to a gas.

I hope this helps. Let me know if anything is unclear.

Answer:

about 200 calories of heat must be added to 1 gram of ethanol to convert it from a liquid to a gas

Explanation:

Pentane, C5H12, will react as shown below:____C5H12(l) + ____O2 --- ______CO2 +_______ H2O

H20(g)

1. Balance the equation above by putting coefficients into each blank space.

1C5H12+ 8O2----- 5CO2+ 6 H2O (this is what I got)

2. How many atoms of oxygen are in 144 grams of pentane?

step 1: 144g of C5H12*(1mole/72.17)= 1.995moles of C5H12
step 2: 1.995moles of C5H12*(8moles of O2/1mole of C5H12)= 15.962 moles of O2
step 3: 15.962 moles of O2*(6.02*10^23 atoms of O2/1mole of O2)= 9.609*10^24 atoms of O2 (I don't know if I did this right)

3. How many atoms of oxygen are in 192 grams of carbon dioxide gas?

step 1: 192 grams of CO2 *(1mole of CO2/44.01grams of CO2)= 4.363 moles of CO2
step two: 4.36moles of CO2* ( 1 mole of O2/1mole of CO2)=
step 3: 4.363* (6.02*10^23) = 2.627*10^24 atoms of O2 (I don't know if this is correct.

4. A) CIRCLE ONE: True or False:

reactant you begin with fewer grams of is your limiting reactant.

B) If 32 g of oxygen gas and 32 g of pentane are combined, how much water will be produced, in moles? (Use your reaction from the top of the first page)

NEED HELP ASAP THANK YOU!!!

Answers

1) That looks right

2) Step 1: Moles=Mass/Mr, so it's 144/72= 2mol
Step 2: Stoichiometry means the ratio of oxygen to pentane is 1:8 so moles of oxygen is moles of pentane times 8. 2*8=16mol of oxygen
Step 3: 16= 9.6*10^24

3) The ratio of oxygen to carbon in CO2 is 1:2
so it's 192/3 = 64
64*2=128g
Moles = Mass/Mr = 128/32 = 4mol = 2.41*10^24 atoms of oxygen
Basically you needed to work out the mass of oxygen using ratios before using a moles equation instead of just working with moles

4) True

5) 1C5H12+ 8O2----- 5CO2+ 6 H2O
32g 32g

Moles of oxygen = Mass/Mr = 32/32= 1mol so you'd need 0.125mol of pentane to react fully
Moles of pentane= Mass/Mr = 32/72=0.4mol so you'd need 3.2mol of oxygen to react fully

This makes oxygen the limiting factor, because you need more of it to react fully with the pentane, so we use oxygen

Stoichiometry means for every 8mol of oxygen there's 5mol of CO2
So, 1/8*5=0.635mol of CO2

List 5 physical properties of matter. How many atoms of each type of element in C6H12O6?

Answers

Physical properties= Color, density, volume, mass, melting point and boiling point

Atoms of each type- C 6, H 12, O 6, total of 24 atoms in one molecule

What is the mass percent of oxygen in sodium bicarbonate (NaHCO3)?

Answers

The answer is 57.14%.

First we need to calculate molar mass of NaHCO3. Molar mass is mass of 1 mole of a substance. It is the sum of relative atomic masses, which are masses of atoms of the elements.

Relative atomic mass of Na is 22.99 g
Relative atomic mass of H is 1 g
Relative atomic mass of C is 12.01 g
Relative atomic mass of O is 16 g.

Molar mass of NaHCO3 is:
22.99 g + 1 g + 12.01 g + 3 · 16 g = 84 g

Now, mass of oxygen in NaHCO3 is:
3 · 16 g = 48 g

mass percent of oxygen in NaHCO3:
48 g ÷ 84 g · 100% = 57.14%

Therefore, the mass percent of oxygen in sodium bicarbonate is 57.14%.

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