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3 years ago

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 years ago

1. Using the Slater rule, determine the effective nuclear charge of platinum.

AleksandrR [38]

Answer:

Z* = 3.55

Explanation:

Slater rule says that:

Z*= Z - S

Z* be the nuclear effective charge

Z is the nuclear charge

S is the shielding constant

First we write the electronic configuration of platinum: $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 4f^{14} 5d^{9} 6s^{1}$

The first Slater rule says that we need to group:

$(1s^{2}) (2s, 2p)^{8} (3s, 3p)^{8} (3d^{10}) (4s, 4p)^{8} (4d^{10}) (5s, 5p)^{8} (4f^{14}) (5d^{9}) (6s^{1})$

The second rule says that the electrons to the right are not shielding, but we are going to solve the exercise for the last level (6s), so we don't have electrons to the right.

For the third rule we have two considerations, if is ns or np and if is nd or nf:

For our case, we have an electro that is in ns, so the rule says that

-electrons within same group shield 0.35, except the 1s which shield 0.30

-electrons within the n-1 group shield 0.85

-electrons within the n-2 or lower groups shield 1.00

Now we can proceed with the calculation:

The first consideration in the third rule does not apply as we only have one electron on this level.

The second consideration will be as follow for the level 5, where we have 17 electrons.

Finally the third consideration will be for levels 1, 2, 3 and 4, where we have 14 for 4f, 10 for 4d, 8 for 4s and 4p, 10 for 3d, 8 for 3s and 3p, 8 for 2s and 2p and finally 2 for 1s, which gives 60 electrons.

So the result for S=(60*1.00 + 17*0.85) = 74.45

And the equation is: Z* = 78 - 74.45

So Z* = 3.55

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