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IgorLugansk [536]
3 years ago
9

please help asap will give brainiest What is the independent variable in an experiment? A) The variable that remains the same th

roughout the experiment. B) The variable being tested in the experiment. C) The variable being measured in an experiment.
Chemistry
1 answer:
iogann1982 [59]3 years ago
5 0
A because if it remains through the whole expídeme y the same the it doesn’t depend on anything else to change it and I got it correct
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Add the polynomials (7x^3-2x^2-12)+(3x^3-8x^2+10x)
laiz [17]

Answer:

10 x ^3 − 10 x ^2 + 10 x − 12

Explanation:

4 0
3 years ago
Using the molecular weight (258 g/mol) determine the amount of mmol of (3S)-2,2,- dibromo-3,4-dimethylpentane required.
Varvara68 [4.7K]

The question is incomplete, the complete question is;

With all of this data in hand, we can now set up our reaction. To begin we are going to use 0.7 g of (35)-2.2-dibromo-3,4-dimethylpentane. Question #7: Using the molecular weight (258 g/mol) determine the amount of mmol of (3S)-2,2,- dibromo-3,4-dimethylpentane required. Round to the tenths place

Answer:

2.70 mmols

Explanation:

Given that;

Mass of (3S)-2,2,- dibromo-3,4-dimethylpentane = 0.7 g

Molar mass of (3S)-2,2,- dibromo-3,4-dimethylpentane = 258 g/mol

From,

Number of moles = mass/molar mass

Number of moles of (3S)-2,2,- dibromo-3,4-dimethylpentane = 0.7g/258 g/mol = 2.7 ×10^-3 moles

Therefore;

Number of moles of (3S)-2,2,- dibromo-3,4-dimethylpentane = 2.70 mmols

8 0
4 years ago
400 ml of a 75 M solution of H2SO4 is needed to for a lab. The stock solution is 16.0 M. Calculate how much stock is needed to m
Len [333]

Answer:

The volume of stock solution needed to make the solution is 1875 ml

Explanation:

The parameters given are;

The volume of 75 M solution of H₂SO₄ = 400 ml

The concentration of stock solution = 16.0 M

Number moles per liter of stock solution = 16 moles

Number of moles in required 400 ml solution = 0.4×75 = 30 M

Volume of stock solution that contains 30 M = 30/16×1 = 1.875 l

The volume of stock solution that is required = 1875 ml

4 0
3 years ago
WHAT MASS OF 1,1 DICHLOROEHTANE MUST BE MIXED WITH 100G OF 1,1 DICHLOROTETRAFLUOROEHTANE TO GIVE A SOLUTION WITH VAPOR PRESSURE
vaieri [72.5K]

This is an incomplete question.

The complete question is:

1,1-dichlorotetrafluoroethane, CF3CCL2F, has a vapor pressure of 228 torr. What mass of 1,1-dichloroethane must be mixed with 100.0 g of 1,1-dichlorotetrafluoroethane to give a solution with vapor pressure 157 torr at 25 degrees celsius?

Answer: 46.9 g of 1,1 dichloroethane must be fixed with 100 g  of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at 25^0C

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

\frac{p^o-p_s}{p^o}=i\times x_2

where,

\frac{p^o-p_s}{p^o}= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

x_2 = mole fraction of solute  

=\frac{\text {moles of solute}}{\text {total moles}}

Given : x g of solute is present in 100 g of solvent

moles of solute (1,1 DICHLOROEHTANE) = \frac{\text{Given mass}}{\text {Molar mass}}=\frac{xg}{98.96g/mol}moles

moles of solvent (1,1 DICHLOROTETRAFLUOROEHTANE ) = \frac{\text{Given mass}}{\text {Molar mass}}=\frac{100g}{170.92g/mol}=0.58moles

Total moles = moles of solute  + moles of solvent = \frac{xg}{98.96g/mol}+0.58

x_2 = mole fraction of solute   =\frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}

\frac{228-157}{157}=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}

0.45=1\times \frac{\frac{xg}{98.96g/mol}}{\frac{xg}{98.96g/mol}+0.58}

x=46.9g

Thus 46.9 g of 1,1 dichloroethane must be fixed with 100 g  of 1,1 dichlorotetrafluoroethane to give a solution with vapor pressure of 157 torr at 25^0C

5 0
4 years ago
An orbital would never exist in the quantum description of an atom is
Mamont248 [21]
Hydrogen i think. the orbital of hydrogen is peculiar
6 0
3 years ago
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