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Lorico [155]
2 years ago
11

Determine how many ml of water you need to remove, by evaporation, if you have a 500 ml of 10.20 M HNO3 dilute solution and you

need to make it a 12 M HNO3 stock solution.
Chemistry
1 answer:
Ludmilka [50]2 years ago
7 0

The total volume of water that would be removed will be 75 mL

<h3>Dilution equation</h3>

Using the dilution equation:

M1V1 = M2V2

In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M

Substitute:

V2 = 500 x 10.20/12

         = 425 mL

The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:

500 - 425 = 75 mL

More on dilution can be found here: brainly.com/question/7208939

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What is the concentration (M) of CH3OH in a solution prepared by dissolving 34.4 g of CH3OH in sufficient water to give exactly
bonufazy [111]

Answer:

4.67M

Explanation:

The concentration of methanol (CH3OH) can be calculated using the following:

Molarity (M) = number of moles(n)/volume(v)

However, mole is not given. It can be obtained by using:

Mole = mass / molar mass

Where; mass = 34.4g

Molar mass (MM) of CH3OH is:

= 12 + 1(3) + 16 + 1

= 12 + 3 + 17

= 32g/mol

mole = 34.4/32

mole = 1.075mol

The volume needs to be converted to L by dividing by 1000

230mL = 230/1000

= 0.230L

Molarity = mol/volume

Molarity = 1.075/0.230

Molarity = 4.6739

Molarity = 4.67M

The concentration of CH3OH in solution is 4.67M

5 0
2 years ago
What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m
bulgar [2K]

Answer:

T_f for given question is 2.79 and T_b is 0.52

\Delta T_b = I \times K_b \times m {i- vant hoff’s constant ; Kb- constant ; m molarity }

M = no. of moles of the solute present in one kg of solution

Let the weight of amount of solute be “w” and its molecular mass be “M”

Let the mass of the solvent in the given question be “x”

\Delta T_b = I \times K_b \times (w/M)/ x

\Delta T_b = I \times K_b \times w/Mx

\Delta T_b = 1 \times 0.51 \times1.5/(0.250 \times 58.44) = 0.052

\Delta T_f = M \times K_f = 1.86 \times 1.5 = 2.79

4 0
3 years ago
How many moles of glucose are in 19.1g of glucose?​
Crank

Answer:

0.106 mol (3s.f.)

Explanation:

To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.

Ar of C= 12, Ar of H= 1, Mr of O= 16

These Ar values can be found on the periodic table.

Mr of glucose= 6(12)+ 12(1) + 6(16)= 180

Moles of glucose

= mass ÷ mr

= 19.1 ÷ 180

= 0.106 mol (3 s.f.)

3 0
3 years ago
If Salt is NaCl, what atoms does the compound contain?
allochka39001 [22]
Na is the elemental abbreviation for sodium, and Cl is the abbreviation for chlorine.
When the two are combined, you get sodium-chloride, or table salt.
Hope that helped =)
8 0
2 years ago
What does the roman numeral stand for in copper(1) oxide should it not be copper(II) oxide
photoshop1234 [79]

Answer:

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

Explanation:

Roman numeral is used to indicate the oxidation number of an element in a compound.

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

This can be seen from the following illustration:

copper(I) oxide => Cu₂O

Oxidation number of O = –2

Oxidation number of Cu₂O = 0

Oxidation number of Cu =?

Cu₂O = 0

2Cu + O = 0

2Cu – 2 = 0

Collect like terms

2Cu = 0 + 2

2Cu = 2

Divide both side by 2

Cu = 2/2

Cu = 1

Thus, we can see that the oxidation number of Cu in Cu₂O is 1. Hence the name of Cu₂O is copper(I) oxide indicating that the oxidation number of of copper (Cu) in the compound is 1.

For copper(II) oxide, we shall determine the oxidation number of Cu. This can be obtained as follow:

copper(II) oxide, CuO => CuO

Oxidation number of O = –2

Oxidation number of CuO = 0

Oxidation number of Cu =?

CuO = 0

Cu + O = 0

Cu – 2 = 0

Collect like terms

Cu = 0 + 2

Cu = 2

Thus, the oxidation number of Cu in CuO is 2. Hence the name of CuO is copper(II) oxide indicating that the oxidation number of of copper (Cu) in the compound is 2.

From the above illustrations,

We can see that the roman numeral in both copper(I) oxide, Cu₂O and copper(II) oxide, CuO are different because the oxidation number of Cu in both cases are different.

3 0
3 years ago
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