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3 years ago

15

Nickel metal is added to a solution of copper(ii) nitrate. express your answer as a chemical equation. identify all of the phase

s in your answer. enter no reaction if no reaction occurs.

1 answer:

seraphim [82]3 years ago

3 0

Nickel is the 28th chemical in the periodic table and the chemical symbol of copper (II) nitrate is Cu(NO₃)₂. The chemical reaction that occurs between these compounds is a single-replacement reaction.

From the activity series of metals, nickel is able to replace copper in the compound and this is expressed in the balanced equation,

Ni(s) + Cu(NO₃)₂(aq) --> Cu(s) + Ni(NO₃)₂(aq)

The phases are also written in the equation.

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aivan3 [116]

Table salt is a compound because Sodium (Na) and Chlorine (Cl) were combined to make a new group.

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2 years ago

The half-life of plutonium-239 is 24300. If a nuclear bomb released 8kg of this isotope, how many years would pass before the am

Aneli [31]

Answer:

= 72900 years

Explanation:

The half-life is the time taken by a radioactive material to decay by half the original amount.
The half-life of plutonium-239 is 24300 years which means it takes 24300 years to decay by half the original amount.

To calculate the time taken for a mass of 8 kg to decay to 1 kg we use;

New mass = Original mass x (1/2) ^n, where n is the number of half-lives

Therefore;

1 kg = 8 kg × (1/2)^n

1/8 = (1/2)^n

solving for n;

n =3

Therefore;

Time = 3 × 24300 years

= 72900 years

It will, therefore, take 72900 years for 8 kg of plutonium-239 to decay to 1 kg.

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3 years ago

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erastovalidia [21]

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2 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:

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

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

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

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