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solniwko [45]
3 years ago
6

If the density of a certain spherical atomic nucleus is 1.0 _ 1014 g cm_3 and its mass is 2.0 _ 10_23 g, what is its radius in c

m? I need to understand how to do it not only the answer please :)
Chemistry
1 answer:
melamori03 [73]3 years ago
5 0
1) Calculate the volume from d = m/V => V = m/d = 2.0*10^-23 g /  1.0*10^14 g/cm^3 = 2.0*10^-9 cm^3

2) Now use the formula of volume for a sphere: V = (4/3)π(r^3) =>

r =∛[3V/(4π)] = ∛[(3*2.0*10^-9 cm^3) / (4π)] = 0.48*10^-3 cm = 4.8*10^-4 cm = 0.00048cm
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The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.
Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0
3 years ago
Which statement best descibes what happens when a person sees a orange t-shirt?
tangare [24]
B. light reflects off of the t shirt and enters the human eye
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3 years ago
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Of these non-metals which one is likely to be least reactive
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Answer:

the answer is the swecond option

Explanation:

Its b ur well come

7 0
3 years ago
Moseley made revisions to the periodic table that resolved some of the problems with Mendeleev’s version. Which of these was a r
olga55 [171]
The result of Moseley's revisions were that the elements were arranged in atomic number order rather than atomic mass order.
5 0
3 years ago
How many grams of propane are in 20 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three
Dima020 [189]

Answer:

a. =9.1x10^3gC_3H_8

b. 2.1x10^2molC_3H_8

c. Q=-4.6x10^5kJ

Explanation:

Hello,

a. By applying the given information, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8} =9.1x10^3gC_3H_8

b. By knowing that the propane has a molecular mass of 44g/mol, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8}*\frac{1molC_3H_8}{44gC_3H_8} =2.1x10^2molC_3H_8

c. Here, the propane's combustion chemical reaction is stated:

C_3H_8+5O_2-->3CO_2+4H_2O

This enthalpy of reaction is computed via:

ΔrH=(3*-393.5kJ/mol)+(4*241kJ/mol)-(-104.7kJ/mol)=-2043kJ/mol

Finally, since it is done for 20 lb of propane (2.1x10^2molC_3H_8), the obtainable energy is:

Q=-2043kJ/mol*2.1x10^2molC_3H_8\\Q=-4.6x10^5kJ

Best regards.

4 0
3 years ago
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