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Brilliant_brown [7]
3 years ago
9

How do the test variables (independent variables) and outcome variables (dependent variables) in an experiment compare? A. The t

est variables (independent variables) and outcome variables (dependent variables) are the same things. B. The test variable (independent variable) controls the outcome variable (dependent variable). C. The test variable (independent variable) and outcome variable (dependent variable) have no affect on each other. D. The outcome variable (dependent variable) controls the test variable (independent variable).
Chemistry
2 answers:
viktelen [127]3 years ago
7 0

Answer:

The test variable (independent variable) controls the outcome variable (dependent variable)

Explanation:

its right on study island

Ghella [55]3 years ago
6 0

Answer:

I'm on the exact same queston

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the density of aluminum is 2.70g/cm^3. a piece of aluminum foil has a volume of 54.0 cm^3. what is the mass of this piece of alu
Neporo4naja [7]
The  mass  of  aluminium  foil   is   calculated   as  follows
mass  =  density  x  volume
density =  2.70  g/cm^3
volume  54  cm^3
mass  of   aluminium  foil  is  therefore  =  2.70  g/cm^3  x  54  cm^3  =145.8  grams
cm^3  cancel   out  each   other
6 0
3 years ago
Balance the combustion eqation; __C5H + __O2 ---> __CO2+__H2O
stira [4]
This equation C5H + O2 ---> CO2 + H2O has a mistake.

C5H is wrong. You missed the subscript of H.

I will do it for you assuming some subscript to show you the procedure, but you have to use the right equation to get the right balanced equation.

Assuming the tha combustion equation is C5H12 + O2 ---> CO2 + H2O

First you need to  balance C, so you put a 5 before CO2 and get

C5H12 + O2 ---> 5CO2 + H2O

Now you count the hydrogens: 12 on the left and 2 on the right. So put a 6 before H2O and get:

C5H12 + O2 ---> 5CO2 + 6H2O

Now count the oxygens: 2 on the left and 16 on the right, so put an 8 on before O2:

=> C5H12 + 8O2 ---> 5CO2 + 6H2O.

You can verify that the equation is balanced
8 0
2 years ago
At which temperature and pressure will a sample of neon gas behave most like an ideal gas?
Andreas93 [3]

Answer:

At STP, 760mmHg or 1 atm and OK or 273 degrees celcius

Explanation:

The standard temperature and pressure is the temperature and pressure at which we have the molecules of a gas behaving as an ideal gas. At this temperature and pressure, it is expected that the gas exhibits some properties that make it behave like an ideal gas.

This temperature and pressure conform some certain properties on a gas molecule which make us say it is behaving like an ideal gas. Ordinarily at other temperatures and pressures, these properties are not obtainable

Take for instance, one mole of a gas at stp occupies a volume of 22.4L. This particular volume is not obtainable at other temperatures and pressures but at this particular temperature and pressure. One mole of a gas will occupy this said volume no matter its molar mass and constituent elements. This is because at this temperature and pressure, the gas is expected to behave like an ideal gas and thus exhibit the characteristics which are expected of an ideal gas

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They are eukaryotic, which means they have a nucleus. Most have mitochondria

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Option B is your answer.
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