All categories

2 years ago

How many grams of C8H18 is in 405,576.398 mL

Chemistry

1 answer:

nevsk [136]2 years ago

4 0

$Molarity X \frac{0.762 mol}{05.576398 L} = Mass$ is the of $C_8H_1_8$ is in 405,576.398 mL.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Given data:

Molar mass of $C_8H_1_8$ = 0.762 mol

Volume of solution = 405,576.398 mL = 405,576.398 mL/1000 = 405.576398 L

Solution:

Step 1: Calculating moles of $C_8H_1_8$

• Molarity = $\frac{\;Moles \;of \;solute}{\;Volume \;of \;solution \;in \;litre}$

• Putting the value in the given formula

• Moles of solute C8H18 = $\frac{\;Mass}{\;Molar \;molar \;mass}$

• Moles of solute C8H18 = $\frac{\;Mass}{0.762 mol}$

Now,

Molarity = $\frac{\;Moles \;of \;solute}{\;Volume \;of \;solution \;in \;litre}$

Molarity = $\frac{\frac{\;Mass}{0.762 mol}}{\;Volume \;of \;solution \;in \;litre}$

Molarity = $\frac{\frac{\;Mass}{0.762 mol}}{405.576398 L}$

$Molarity X \frac{0.762 mol}{05.576398 L} = Mass$

Hence, $Molarity X \frac{0.762 mol}{05.576398 L} = Mass$ is the of $C_8H_1_8$ is in 405,576.398 mL.

Learn more about molarity here:

brainly.com/question/2817451

#SPJ1

You might be interested in

Which statements about the equation shown above are correct? Check all that apply. It represents a reaction that involves rearra

mezya [45]

Answer: a and c

Explanation:

6 0

2 years ago

Read 2 more answers

When your heart doesn't know what it wants.. what do you do?

DENIUS [597]

Follow your nose to the fruity taste of chows recharge the heart then follow your heart good luck my friend

3 0

3 years ago

i'm reasking this, i need help with this and it's due today. the answer isn't 30,240 minutes, it's the blanks i need filled in :

adelina 88 [10]

Answer:

Explanation:

24hours ×21 days=

you will get the answeer

3 0

3 years ago

You are given a solution that is 518 mM lactose. You need to make up 4.5 L of 16.7 mM solution. What volume do you need to trans

OLga [1]

Answer:

The volume you need to transfer from the stock solution is 0.145 l

Explanation:

Since the number of moles of lactose in the volume of stock solution that you transfer will be the same as the number of moles of lactose in the final solution, you can use this expression:

number of moles in volume to transfer = number of moles in the final solution

Since number of moles = concentration * volume (if the concentration is expressed in molarity), then:

Ci * Vi = Cf * Vf

where:

Ci = concentration of the stock solution.

Vi = volume of the stock solution to be transferred.

Cf = concentration of the final solution

Vf = volume of the final solution

Then, replacing with the data:

518 mM * Vi = 16.7 mM * 4.5 l

Vi = 16.7 mM * 4.5 l / 518 mM

Notice that any concentration unit can be used, as long as the units of the concentration of the stock and final solution are the same.

4 0

3 years ago

Calculate Ho298 for the process

Inga [223]

Explanation:

As per the 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.

Hence, 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. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}$ $\Delta H^0_1 = -314 kJ$ ..........(1)

$SbCl_{3} + Cl_2 \rightarrow SbCl_{5}$ $\Delta H^0_2 = -80kJ$ ..............(2)

The final reaction is as follows:

$Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}$ $\Delta H^0_3 = ?$ .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of $\Delta H^{0}_{3}$ at 298 K will be as follows.

$\Delta H^{0}_{3}$ = $\Delta H^{0}_{1}$ + $\Delta H^{0}_{2}$

= -314 kJ + (-80) kJ

= -394 kJ

Thus, we can conclude that $H^{o}$ at 298 K for the given process is -394 kJ.

4 0

3 years ago