The state of Oregon has an entire area of 9.70 x 10^4

1. What volume of toluene do you need to add to 1 mL of ethyl acetate to make an equi-molar mixture? The density of ethyl acetat

galben [10]

Answer:

1.06 mL of toluene will be needed.

Explanation:

Equi-molar mixture means equal moles of all the components.

as given the volume of ethyl acetate = 1mL

Density of ethyl acetate = 0.898 g/mL

The relation between density, mass and volume is :

$density=\frac{mass}{volume}$

mass=volumeXdensity

mass of ethyl acetate present = 1mL X 0.898g/mL = 0.598 grams

the moles are related to mass as:

$moles=\frac{mass}{molarmass}$

For ethyl acetate molar mass = 4X12+8X1+2X16= 88g/mol

moles of ethyl acetate will be:

$moles=\frac{0.898}{88}= 0.01mol$

So we need 0.01 moles of toluene also

For 0.01 moles the mass of toluene required = 0.01 X molar mass of toluene

mass required = 0.01 X 92=0.92grams

for 0.92 grams of toluene volume required will be:

$volume=\frac{mass}{density}=\frac{0.92}{0.867}= 1.06mL$

4 0

2 years ago

What are the 3 critical components of an electromagnet and what purpose do they each serve?

WARRIOR [948]

Answer:

A source of electricity, a wire coil, and an iron core

Explanation:

An electromagnet has three critical components:

1. A source of electricity

This is often a battery.

It generates the electric current that produces the magnetic field.

2. A wire coil

The wire carries the electric current.

Stacking the wire into loops makes a stronger magnetic field.

The more loops in the coil, the stronger the field.

3. An iron core

An iron core greatly increases the strength of the magnetic field within it and at its ends.

4 0

2 years ago

Determine the normality of the following solutions note the species of interest is H 95 g of PO4 3- in 100mL solution

Aliun [14]

Answer : The normality of the solution is, 30.006 N

Explanation :

Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.

Mathematical expression of normality is:

$\text{Normality}=\frac{\text{Gram equivalent of solute}}{\text{Volume of solution in liter}}$

or,

$\text{Normality}=\frac{\text{Weight of solute}}{\text{Equivalent weight of solute}\times \text{Volume of solution in liter}}$

First we have to calculate the equivalent weight of solute.

Molar mass of solute $PO_4^{3-}$ = 94.97 g/mole

$\text{Equivalent weight of solute}=\frac{\text{Molar mass of solute}}{\text{charge of the ion}}=\frac{94.97}{3}=31.66g.eq$

Now we have to calculate the normality of solution.

$\text{Normality}=\frac{95g}{31.66g.eq\times 0.1L}=30.006eq/L$

Therefore, the normality of the solution is, 30.006 N

5 0

2 years ago

Marie mixed 5g of carbon with 5g of lead oxide. she heated the mixture strongly for 15 minutes in a fume cupboard.

olga2289 [7]

The problem talks about two questions and these are:

1. Metals are very good conductors of electricity and heat. Directing heat is easier. So let Marie heat the beads and also have heat another substance, for instance, water. If the beads heat quicker, then they are metals. Another test to conduct is called flame test. This test should give you a colored flame (blue/white for lead) the metal is lead if the reaction is: 2PbO+C ==> 2Pb +CO2

2. The beads are possibly to be lead since Ferrous(lead) oxide + carbon = carbon dioxide + lead

7 0

2 years ago

Coal gasification is a multistep process to convert coal into cleaner-burning fuels. In one step, a coal sample reacts with supe

ddd [48]

Answer :

The enthalpy of reaction is, -187.6 kJ/mol

The total heat will be, -2251 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

(a) The formation of $CH_4$ will be,