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oksano4ka [1.4K]
2 years ago
9

a researcher obtains a sample of 0.070 M nitrate solition. A 20.0 mL aliquote of the nitrate solution is added to 10.0 mL of amm

onia and diluted to 100 ml with deionized water. what is the concentration of nitrate in the new solution
Chemistry
1 answer:
katovenus [111]2 years ago
6 0

Answer:

Concentration of nitrate in the new solution = 0.007 M

Explanation:

Given:

Concentration nitrate solution = 0.070 m

Volume of aliquote of the nitrate solution is add = 10.0 ml

Total volume = 100 ml

Find:

Concentration of nitrate in the new solution

Computation:

Number of M. mole = 0.070 m x 10.0 ml

Number of M. mole = 0.7 m-moles

Concentration of nitrate in the new solution = 0.7 m-moles / 100 ml

Concentration of nitrate in the new solution = 0.007 M

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Illusion [34]

Answer:

6He =   4.90 MeV/Nucleon

8Li =     5.18  MeV / Nucleon

62Ni =   8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

Explanation:

Our strategy here is to remember that when the mass of a given nuclei is calculated from the sum of the mass of its  protons and neutrons, this mass is greater than  the actual  value . This is the mass defect.

Now this mass defect we can convert to energy  by utilizing Einstein´s equation, E = mc².This is  the binding energy.

For 6He with actual mass 6.0189 u ( He has Z = 2, that is 2 protons )

mass protons =   2 x 1.0078 u  =  2.0516 u

mass neutrons = 4 x 1.0087 u  =  4.0348 u

predicted mass = (2.0516 + 4.0348) u = 6.0504 u

mass defect = (6.0504 - 6.0189) u = 0.0315 u

Now we need to convert this mass expressed in atomic mass units to kilograms ( 1 u = 1.66054 x 10⁻²⁷ Kg )

0.0315 u x 1.66054 x 10⁻²⁷ Kg =5.231 x 10⁻²⁹ Kg

E =5.231x 10⁻²⁸ Kg x (3 x 10⁸ m/s )² = 4.707 x 10⁻¹²J

Finally we will convert this energy in Joules to eV

E = 4.707 x 10⁻¹²  J x 6.242 x 10¹⁸ eV/J = 2.94 x 10⁷ eV = 29.4 MeV

E per nucleon for 6He = 29.4 MeV / 6 =4.90 MeV / Nucleon

Now the calculations for the rest of the nuclei are performed in similar manner with the following results:

8Li = 5.18 MeV / Nucleon

62Ni = 8.82 MeV / Nucleon

115 In =  8.54 MeV / Nucleon

8 0
3 years ago
If 17.8 grams of KOH dissolve in enough water to make a 198-gram solution, what is the concentration in percent by mass?
sleet_krkn [62]
Solute of solution = 17.8 g

Solvn = 198 g

% = 17.8 / 198

w% = 0.089 x 100 = 8.9%  by mass

hope this helps!
7 0
3 years ago
Molar mass of Ag2SO4<br><br> Full work Pls
IceJOKER [234]
This would be the molar mass.

5 0
3 years ago
What volume of 6.58M HCI is needed to make 500. mL of 3.00M HCI?
zmey [24]

Answer:

228 mL

Explanation:

M1*V1 = M2*V2

M1 = 6.58 M

V1 = ?

M2 = 3.00 M

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3 0
3 years ago
Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov
love history [14]

Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field  = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3

Mass of the Astroturf = m

Density of the Astroturf = d = 187 oz/ft^3

d=\frac{m}{V}

m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz

1 oz = 0.0283495 kg

m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg

Weight of the Astroturf = W

W = mg

=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N

The weight of the Astroturf is 179,684.31 Newtons.

8 0
3 years ago
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