How many milliliters of 60% carbonic acid must be mixed with how many milliliters of 15% carbonic acid to make 650 milliliters of a 38% carbonic acid solution

Answers

Answer 1

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:

650 mL * 38/100 = 247 mL

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

1)  x * 60/100 + y * 15/100 = 247

for the volume of carbonic acid; and

2)  x + y = 600 mL

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):

x = 600 - y

And replace x in equation 1):

(600 - y) * 60/100 + y * 15/100 = 247360 - 0.6y + 0.15y = 247-0.45y = -113y = 251. 1 mL

Finally we calculate x using equation 2):

x + 251.1 = 600x = 348.9 mL


Related Questions

Give an example of chemical reaction that occurs in everyday life

Answers

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! :)

An isotope has three forms. 30% have a mass of 4 amu, 20% have a mass of 5 amu and
50% have a mass of 3 amu, Average atomic mass will be closest to

Answers

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

Give the IUPAC name for each compound.

Answers

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).

Best regards!

What carpet Burns in a deficiency of O2 a mixture of CO and CO2 forms.Carbon Burns in excess O2 to form only CO2 and CO Burns in excess O2 to form only CO2. Calculate ΔH for C(graphite +1/2O2) →CO(g).

Answers

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

What is a molecule contains the genetic instructions for the development and functioning of all living organisms found in the nucleus of a cell

Answers

DNA (short for deoxyribonucleic acid)

Which word describes the amount of matter an object contains?
altitude
density
mass
pressure

Answers

Mass witch would be c i covered this in class

Answer:

Mass

Explanation:

Answered this question multiple times and got it correct

Hope this helped!

Stay safe!

(1) Phosphorus silicon is a semi-conductor

Answers

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:

A patient is prescribed 100mg/day of antibiotic for 2 weeks. The antibiotic is available in vials that contain 20mg/vial of the drug. How many vials are necessary for the entire treatment?

Answers

i have no idea, but i’m pretty sure you gotta divide

CHEMISTRY HELP
Once the following equation is balanced, what is the correct coefficient for Z₂?

Answers

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.

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