Empirical formula for 70.9% K and 29.1% S

A chemist prepares a solution of silver perchlorate by measuring out of silver perchlorate into a volumetric flask and filling t

djverab [1.8K]

Complete Question:

A chemist prepares a solution of silver (I) perchlorate (AgCIO4) by measuring out 134.g of silver (I) perchlorate into a 50.ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the silver (I) perchlorate solution. Round your answer to 2 significant digits.

Answer:

13 mol/L

Explanation:

The concentration in mol/L is the molarity of the solution and indicates how much moles have in 1 L of it. So, the molarity (M) is the number of moles (n) divided by the volume (V) in L:

M = n/V

The number of moles is the mass (m) divided by the molar mass (MM). The molar mass of silver(I) perchlorate is 207.319 g/mol, so:

n = 134/207.319

n = 0.646 mol

So, for a volume of 50 mL (0.05 L), the concentration is:

M = 0.646/0.05

M = 12.92 mol/L

Rounded to 2 significant digits, M = 13 mol/L

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

A preliminary untested explanation that tries to explain how or why things happen in the manner observed is a scientific

icang [17]

It is a scientific hypothesis. A scientific hypothesis must be testable, however there is a significantly more grounded necessity that a testable speculation must meet before it can truly be viewed as logical. This foundation comes essentially from crafted by the rationalist of science Karl Popper, and is called "falsifiability".

3 0

4 years ago

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Unknown element x is a metal that ionically bonds to sulfur.is the formula, x3s feasible? why or why not?a)it is feasible. the t

nasty-shy [4]

The correct answer is:
d) No, it is not feasible. three metallic ions cannot provide the exact number of electrons that one sulfur needs for the ionic bond.
Because Sulfur is divalent so it need to gain 2 electrons from metal so if we have 3 metals they can't provide only two electrons only.

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3 years ago

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Kendra and Jordan gather firewood. The wood is cool to the touch, but the campfire they make warms their hands. Which statement

Free_Kalibri [48]

Answer:

D

Explanation:

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3 years ago