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shepuryov [24]
2 years ago
6

You have run an experiment studying the effects of the molecular weight of a compound on the rate of diffusion in agar. Compound

X has a molecular weight of 25.3 g/mol and compound Y has a molecular mass of 156.2 g/mol. On two separate agar plates, 0.1 g of each substance were transferred and allowed to diffuse for 2 hours. The results below were obtained. Use this information about this situation to answer the following three questions.
a. What was the independent variable in this experiment?
b. What was the independent variable in this experiment?
c. List two controls that were held constant for this experiment or that you would hold constant for this experiment
Chemistry
1 answer:
Naddika [18.5K]2 years ago
7 0

Answer:

A) Molecular weight

B) Rate of diffusion

C) Agar plates

Time of diffusion

Explanation:

A) The independent variable is the molecular weight because it is the variable that the researcher changes at will to aid his research. It doesn't depend on other variables.

B) The dependent variable is "Rate of diffusion because it depends on the molecular weight of the compound.

C) A control variable is one that would be held constant throughout the research.

In this case, the agar plates and the time of 2 hours diffusion remain the same throughout.

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Answer:

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Classify each item by matching as organic
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sucrose is composed of carbon hydrogen and oxygen which of the following is not needed in order to determine the precent by mass
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It is often desirable to determine the mass percent of elements in a given compound.

To determine the mass percent of elements:

  • Evaluate the formula mass of the compound. This is done by summing the atomic masses of the atoms in the compound together.
  • The mass percentage is determined by pacing the mass contribution of each element or group to the formula mass of the compound and multiply by 100.

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3 0
3 years ago
Hat happens when an atom gains an electron?
Aliun [14]
A) It becomes a negative ion
7 0
3 years ago
A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.
Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

3 0
3 years ago
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