Using the periodic table, choose the more reactive metal. Ta or V Ta

Calculate the percentage composition of oxygen in aluminium nitrate, Al(NO3)3.

kotegsom [21]

Answer:

A small still is separating propane and butane at 135 °C, and initially contains 10 kg moles of a mixture whose composition is x = 0.3 (x = mole fraction butane). Additional mixture (x = 0.3) is fed at the rate of 5 kg mole/hr. The total volume of the liquid in the still is constant, and the concentration of the vapor from the still (xp) is related to x, as follows: Xp = How long will it take for X, to change from 0.3 to 0.35.

7 0

2 years ago

Which reactant (X) is missing from the equation shown? X+ PO4 → H3PO4

insens350 [35]

Answer:

H

Explanation:

b/c the polyatomic PO4¯³

have 3 valance number

3 0

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:

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

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

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

7 0

2 years ago

What amount (moles) of compound is present in 1.00 g of each of the compounds in Exercise 54?

Anika [276]

Answer:

1. Caffeine, C₈H₁₀N₄O₂

Amount = 1.00/194 = 0.00515 moles

2. Ethanol, C₂H₅OH

Amount = 0.0217 moles

3. Dry Ice, CO₂

amount = 0.0227 moles

Note: The question is incomplete. The compound are as follows:

1. Caffeine, C₈H₁₀N₄O₂;

2. Ethanol, C₂H₅OH;

3. Dry Ice, CO₂

Explanation:

Amount (moles) = mass in grams /molar mass in grams per mole

1. Caffeine, C₈H₁₀N₄O₂

molar mass of caffeine = 194 g/mol

Amount = 1.00 g/194 g/mol = 0.00515 moles

2. Ethanol, C₂H₅OH

molar mass of ethanol = 46 g/mol

Amount = 1.00 g/46 g/mol = 0.0217 moles

3. Dry Ice, CO₂

molar mass of dry ice = 44 g/mol

amount = 1.00 g/44 g/mol = 0.0227 moles

5 0

2 years ago

Calculate the number of moles of each compound, given the number of molecules.

brilliants [131]

Answer:

1. $2.46\times 10^{21}$ molecules of CO₂

2. 10⁴ molecules of H₂O

3. 8.75×10³² molecules of C₆H₁₂O₆

Explanation:

1. $2.46\times 10^{21}$ molecules of CO₂

2. 10⁴ molecules of H₂O

3. 8.75×10³² molecules of C₆H₁₂O₆

5 0

2 years ago

Using the periodic table, choose the more reactive metal. Ta or V Ta​

Using the periodic table, choose the more reactive metal. Ta or V Ta