A solution is made by mixing 49.g of chloroform CHCl3 and 73.g of acetyl bromide CH3COBr. Calculate the mole fraction of chloroform in this solution. Round your answer to 2 significant digits.

Answer:

The coefficient of Z₂ is 1.

Explanation:

From the question given above:

X + ZY —> XY + Z₂

Next, we shall balance the equation to obtain the coefficient of Z₂. This can be obtained as follow:

X + ZY —> XY + Z₂

There is 1 atom of Z on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of ZY as shown below:

X + 2ZY —> XY + Z₂

There are 2 atoms of Y on the left side and 1 atom on the right side. It can be balance by putting 2 in front of XY as shown below:

X + 2ZY —> 2XY + Z₂

Now, we have 1 atom of X on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of X as shown below:

2X + 2ZY —> 2XY + Z₂

Now the equation is balanced.

Thus, the coefficient of Z₂ is 1.

When Silicone is altered with Phosphorus, 4 of its electrons form covalent bonding and the 5th electron becomes delocalized and thus it increases electrical conduction

Answer:

blue blah blue blah

Explanation:

Answer: metal and salt water

Explanation: salt starts to brake down metal and metal is everywhere in the ocean hope this helps

Answer:

baking a cake.

Explanation:

baking a cake is an example of a chemical reaction that happens daily because a new substance is forming when the cake is being baked. hope this helps! :)

Answer:

348.9 mL of the 60% solution and 251.1 mL of the 15% solution.

Explanation:

First, we calculate how many mililiters of pure carbonic acid are there in 650 mL of a 38% solution:

Then we can express the sum of both initial solutions as:

for the volume of carbonic acid; and

For the volume of the solutions.

We now have a system of two equations and two unknowns (x is the volume of the 60% solution and y is the volume of the 15% solution).

We express x in terms of y in equation 2):

And replace x in equation 1):

Finally we calculate x using equation 2):

Answer:

Explanation:

From the combustion of carbon, the reactions occurring in limited oxygen conditions are:

[tex]C(graphite) + \dfrac{1}{2}O_{2(g)} \to CO_{(g)}[/tex]  

[tex]C(graphite) + O_{2(g)} \to CO_{2(g)}[/tex]  

If it occurs in excess, then any leftover CO changes to CO2. i.e.

[tex]C(graphite) + O_{2(g)} \to CO_{2(g)}[/tex]   ---- (1)

[tex]CO_{(g)} + \dfrac{1}{2}O_{(g)} \to CO_{2(g)}[/tex]          ----- (2)

From (1), the enthalpy change is:

[tex]\Delta H_{rxn1} = \Delta H^0_{fCO_2(g)} - ( \Delta H^0_{f C(graphite)}+ \Delta H^0_{fCO_2(g)}[/tex]

[tex]\Delta H_{rxn1} =-393.5 \ kJ/mol -(0+0)[/tex]

[tex]\Delta H_{rxn1} =-393.5 \ kJ/mol[/tex]

From (2), the enthalpy change is:

[tex]\Delta_{rxn2} = \Delta H^0_{fCO_2(g)} - ( \Delta H^0_{fCO(g)} + \dfrac{1}{2} \Delta H^0_{fO_2(g)})[/tex]

[tex]\Delta_{rxn2} = -393.5 \ kJ/mol -(-110.5 + \dfrac{1}{2}(0))[/tex]

[tex]\Delta_{rxn2} = -283.0 \ kJ/mol[/tex]

Subtracting (2) from (1), we get:

[tex]C(graphite) + O_{2(g)} \to CO_{2(g)} \ \ \ \Delta H_{rxn} = -393.5 \ kJ/mol}[/tex]

[tex]CO_{(g)} + \dfrac{1}{2} O_2(g) \to CO_{2(g)}} \ \ \ \Delta H _{rxn2} = -283.0 \ kJ/mol[/tex]

[tex]C(graphite) + O_{2(g)} \to CO (g) + \dfrac{1}{2}O_{2(g)} \ \ \ \Delta H_{rxn} = -110.5 \ kJ/mol[/tex]

[tex]C(graphite) + \dfrac{1}{2} O_{2(g)} \to CO (g) \ \ \ \Delta H_{rxn} = -110.5 \ kJ/mol[/tex]

The enthalpy change ΔH of the reaction = -110.5 kJ/mol

The average atomic mass will be 3.70 amu

We have to obtain the average atomic mass by multiplying the mass of each isotope in amu by the relative abundance of the given isotope as contained in the data of the question.

Hence;

Average atomic mass;

(0.30 * 4) + (0.20 * 5) + (0.50 * 3)

= 1.20 + 1.00 + 1.50

= 3.70 amu

https://brainly.com/question/13292428

Answer:

Mass

Explanation:

Answered this question multiple times and got it correct

Hope this helped!

Stay safe!

Answer:

a. 1-fluoro-3,3,4-trimethyl-pentane.

b. 1-iodo-3-ethyl-2-methyl-hexane.

c. 1,3-dichloro-5-dimethyl-hexane.

d. 1-bromo-3-methyl-cyclopentane.

Explanation:

Hello!

In this case, according to the IUPAC rules for the listed alkyl halides, we first need to name the halogens (considering periodic order) then alkyl radicals and finally the parent chain; thus, the names are given below:

a. 1-fluoro-3,3,4-trimethyl-pentane.

b. 1-iodo-3-ethyl-2-methyl-hexane.

c. 1,3-dichloro-5-dimethyl-hexane.

d. 1-bromo-3-methyl-cyclopentane.

e. 5-chloro-1-bromo-1,1,5-trimethyl-pentane (radicals are not clear).

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