Please answer the missing problems and check if they are all right

Please help me with this chemistry question, image attached

ss7ja [257]

average atomic mass of Cu = 63.54

Explanation:

average atomic mass = atomic mass of isotope 1 × abundance of isotope 1 + atomic mass of isotope 2 × abundance of isotope 2 +...+ atomic mass of isotope n × abundance of isotope n

average atomic mass of Cu = atomic mass of Cu-63 × abundance of isotope Cu-63 + atomic mass of isotope Cu-65 × abundance of isotope Cu-65

average atomic mass of Cu = 62.93 × (69.17 / 100) + 64.93 × (30.83 / 100)

average atomic mass of Cu = 63.54

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average atomic mass

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6 0

3 years ago

How many minutes will it take to plate out 4.50 g of Cu from a solution of Cu(NO3)2 (aq) onto the cathode of an electrolytic cel

ipn [44]

Answer:

It will take 28.5 minutes

Explanation:

<u>Step 1: </u>Data given

Mass of Cu = 4.50 grams

8.00 A of current are used

Molar mass of Cu = 63.5 g/mol

Step 2: Calculate time needed

Cu2+ →Electricity → Cu

we notice a flow of 2 electrons ⇒ This means the Faraday constant = 2F

Since Molar mass of Cu is 63.5 g/mol

63.5 grams of Cu is deposited by 2*96500 C

4.50 grams of Cu ((2*96500)/63.5) * 4.50 = 13677.17 C

Q = It

13677.17 = 8t*60 seconds

t = 28.5 minutes

3 0

3 years ago

Which type of star has the greatest temperature on the HR diagram? Question 7 options: Red Supergiants Main Sequence White Dwarf

Sauron [17]

Answer:

Red supergiant

Explanation:

Their core is the hottest.

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7 0

3 years ago

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

7 0

3 years ago

For the reaction below, if the rate of appearance of Br2 is 0.180 M/s, what is the rate of disappearance of

brilliants [131]

Answer:

–0.360 M/s

Explanation:

7 0

3 years ago