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mrs_skeptik [129]
3 years ago
12

An atom of 125Sn has a mass of 124.907785 amu. Calculate the binding energy in MeV per atom. Use the masses: mass of 1H atom = 1

.007825 amu mass of a neutron = 1.008665 amu 1 amu = 931.5 MeV Give your answer to 3 significant figures and DO NOT use E notation. No charity points will be awarded.......
Chemistry
1 answer:
Ksenya-84 [330]3 years ago
3 0

Answer:

1.06×10³ MeV/atom

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Also, atomic number of Sn = 5 0

Thus, the number of protons = 50

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

125 =  50 + Number of neutrons

Number of neutrons = 75

Mass of neutron = 1.008665 amu

Mass of proton = 1.007825 amu

Calculated mass = Number of protons*Mass of proton + Number of neutrons*Mass of neutron

Thus,  

Calculated mass = (50*1.007825 + 75*1.008665) amu = 126.041125 amu

Given mass = 124.907785 amu

<u>Mass defect = Δm = |126.041125 - 124.907785| amu = 1.13334 amu </u>

The conversion of amu to MeV is shown below as:-

1 amu = 931.5 MeV

<u>So, Energy = 1.13334*931.5 MeV/atom = 1.06×10³ MeV/atom</u>

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d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

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A client who weighs 70 kg is receiving a solution of 0.9% sodium chloride (normal saline) 500 ml with dopamine 800 mg at 5 ml/ho
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The mcg/kg/minute the client is receiving is 1.904 mcg/kg/min whose weight is 70 kg.

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