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Romashka [77]
3 years ago
6

1. Which variable is the independent variable and which is the dependent variable? Density vs. ethylene glycol

Chemistry
2 answers:
romanna [79]3 years ago
6 0

Answer:

Density is independent variable and mass is dependent variable.

A. 1.015 g/mL is the density of the ethylene glycol solution.

B. 16.378 mol/L is the molarity of the ethylene glycol solution.

Explanation:

Independent variable is defined as the variable which remain as such and has no effect on the change of another variable. For

Dependent variable is defined as the variable whose value varies with respect to any other variable.

Density is an independent factor as one can determine it by the slope of graph of mass verses volume. Density will remain same on changing the mass of the substance.

Where as mass of ethylene glycol is dependent  variable.

A. Mass of volumetric flask with stopper = 32.6341 g

Mass of volumetric flask, stopper and ethylene glycol =  58.0091 g

Mass of ethylene glycol = 58.0091 g - 32.6341 g =25.3751 g

Volume of the volumetric flask = 25 mL

Density=\frac{Mass}{Volume}

Density of ethylene glycol = D

D=\frac{25.3751 g}{25 mL}=1.015 g/mL

B. Molarity=\frac{Moles}{Volume(L)}

Moles of ethylene glycol ,n = \frac{25.3751 g}{62 g/mol}=0.4093 mol

Volume of the solution = V = 25 mL = 0.025 L

Molarity =\frac{0.4093 mol}{0.025 mL}=16.37 mol/L

AnnyKZ [126]3 years ago
5 0
<span>1. Which variable is the independent variable and which is the dependent variable? Density vs. ethylene glycol
The independent variable would be ethylene glycol and dependent variable would be density.

A. A 25-mL volumetric flask with its stopper has a mass of 32.6341 g. The same flask filled to the line with ethylene glycol (C2H6O2, automotive antifreeze) solution has a mass of 58.0091 g. What is the density of the ethylene glycol solution?

Density = 58.0091 - 32.6341 / .025 = 1015 g/L

B. What is the molarity of the ethylene glycol solution, if the mass of ethylene glycol in the solution is 12.0439 g?

Molarity = 12.0439 ( 1 mol / 62.07 g) / 0.025 = 7.8 M</span>
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2. Calculate the pl of the following amino acids(use their Pka values) a. Arginine b. Glutamic acid of water an c. Asparagine d.
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<h2>♨ANSWER♥</h2>

pl (25*C)

Arginine -----> 10.76

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Asparagine -----> 5.43

Tyrosine -----> 5.63

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5 0
2 years ago
You wish to make a 0.375 M hydroiodic acid solution from a stock solution of 6.00 M hydroiodic acid. How much concentrated acid
cluponka [151]

We need to add 2.09 mL of concentrated acid to obtain 75 mL of 0.335 M HBr solution.

You are performing a dilution of HBr going from a concentration of 12M to 0.335M, and you want to end up with a final volume of 75 ml of the dilute solution.

Consider the dilution formula: M1V1 = M2V2.

The basis behind this formula is that the number of moles of the acid before and after the dilution must remain constant.

M1 = the molarity of the stock solution,

M2 = the molarity of the diluted solution,

V2 = the final volume of the diluted solution.

In this case, we need to determine V1, which is the volume of the stock solution used to prepare the diluted sample. With this knowledge, we can plug our numbers into the equation and we obtain the following:

(12 mol/L)*V1 = (0.335mol/L)*(0.075L).

Keeping in mind that molarity is the moles of a substance in one liter of solution, we will use mol/L instead of M. By doing this, we are reminded that in order to use this equation, we must convert 75 mL into units of liters.

After rearranging the equation and solving for V1, we find that V1 = 0.00209L.

Finally, we must convert back from liters to mL by multiplying the final answer by 1000.

This way we end up with V1 = 2.09 mL.

This means that we need to add 2.09 mL of concentrated acid to obtain 75 mL of 0.335 M HBr solution.

Learn more about molarity here: brainly.com/question/23243759

#SPJ4

3 0
2 years ago
How many grams are in 44.8 liters of nitrogen gas, n2?<br> a) 56g<br> b) 27g<br> c) 36g<br> d) 112g
kkurt [141]

a) 56g

<h3>Calculation:</h3>

At STP,

22.4 L of N₂ = 1 mol

We have given 44.8 L of N₂, therefore,

44.8 L of N₂ = \frac{44.8}{22.4}

                    = 2 mol

We know that,

1 mol of N₂ = 28 g

Hence,

2 mol of N₂ = 28 × 2

                   = 56g

Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.

Learn more about calculation at STP here:

brainly.com/question/9509278

#SPJ4

3 0
2 years ago
Read 2 more answers
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