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barxatty [35]
3 years ago
11

Three beakers contain clear, colorless liquids. one beaker contains pure water, another contains salt water, and another contain

s sugar water. how can you tell which beaker is which? (
Chemistry
1 answer:
tiny-mole [99]3 years ago
8 0
First, we'll identify the beaker containing pure water as follows:
We'll take equal masses from each of the three beakers and measure the mass of each.
We'll then identify the density of each by using the rule : density =mass/volume
Pure water will be the liquid having density equal to 1 gm/cm^3

Then, we'll differentiate between the salt and sugar solution by measuring the conductivity of each solution. Salt solution is a good conductor while solution of sugar is a bad conductor.
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you are on spring break and you won a bunch of balloons at a carnival in the mountains of Georgie. You are now driving back to F
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I think they start to deflate bc of gravity.
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What is the significance of enzymes in maintaining homeostasis in living organisms ?
GalinKa [24]
 Enzymes are needed for metabolic pathways in the body, respiration, digestion and other important life processes. When enzymes function properly, homeostasis is maintained. However, if an enzyme is lacking or has an incorrect shape due to genetic mutation, this can lead to disease within an organism.
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3 years ago
A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol
Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the solution

c_1 = specific heat of calorimeter = 12.1J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m_2 = mass of water or solution = Density\times Volume=1/mL\times 100.0mL=100.0g

\Delta T = change in temperature = T_2-T_1=(26.3-20.2)^oC=6.1^oC

Now put all the given values in the above formula, we get:

q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]

q=2623.61J

Now we have to calculate the enthalpy change for the process.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = Concentration\times Volume=1M\times 0.050L=0.050mole

\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0
3 years ago
(Link; https://www.chemteam.info/SigFigs/SigFigsFable.html)
Anna007 [38]

1. The measurement turned out to be so expensive since the student did not rely on the significant figures to calculate the edge of the cube and approximate it to the next value (2.1 cm possibly) that would allow a simpler construction and therefore its cost was much lower .

2. The student had to take all the significant figures in his calculations, with which he would have:

Volume = \frac{mass}{density}

Volume = \frac{80g}{8.67g/mL}

Volume = 9.2272203 mL

Since the figure to be constructed is a cube, he had to calculate the cube root of the volume to find the value of the edge of the cube:

Edge of the cube = \sqrt[3]{9.2272203 cm^{3} } (Taking into account that cm^{3} is proportional to mL)

Edge of the cube = 2.097443624 cm

Because the cube edge value is so specific, in order to manage his budget, he was able to order a 2.1 cm cube, which would bring the mass up to 80.29287 g, and in the lab reduce one of the faces to the appropriate weight. .

On the other hand, the main thing he had to do was ask how much it would cost to make a cube with those specifications, especially when they mentioned that it would be "expensive" and he only had $50.

The significant figures guarantee the correct operation of a machinery, a gear, a team in general, for which the accuracy will not only be taken to the millimeter, but sometimes microns or much more specific, as in the case of computer components, Therefore, it is very important, if not, and if arbitrary measurements are taken that do not consider significant figures, the components could not function properly, which would cause a loss in time, effort and manpower.

If you want to learn more about exercises with significant figures, you can see the next link: brainly.com/question/11904364?referrer=searchResults

4 0
2 years ago
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