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lead-204 203.973amu, 1.40% abundance. lead-206 205.974 amu, 24.10% abundance. lead-207???, abundance 22.10%. lead-208 207.977, abundance 52.40%. Calculate the isotope mass of lead-207
=207.2 amu (Pb)
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Natural boron is composed of two isotopes: 19.78% boron-10 (atomic mass = 10.0129 amu) and 80.22% boron-11 (atomic mass = 11.0093 amu). What is the average atomic mass of naturally occurring boron?
Average atomic mass=(19.78%*10.0129amu + 80.22%*11.0093amu)/100%=10.8122amu
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Observe the following RXN:
CH4 + O2 --> CO + 2H2O
Carbon has a mass of 12 amu, Hydrogen 1 amu and Oxygen 16 amu.
If 16g of CH4 and 32g of O2 are reacted, how much CO is produced?
The amount of CH4 is n(CH4) = m(CH4)/M(CH4) = 16/16 = 1 mol The amount of O2 is n(O2) = M(O2)/M(O2) = 32/32 = 1 mol According to the reaction...
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Fictitious element X has an a average atomic mass of 254.9 and only 2 isotopes, One of its isotopes has an abundance of 72.00 and a mass of 250.9 amu. What is the atomic mass of the second isotope?
Solution:The average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its natural abundance.Average atomic mass = f1M1 + f2M2 +… + fnMn where...
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What is the mass of iron if 842.05 joules are added and the temperature moves from 25 <sup>o</sup>C to 28 <sup>o</sup>C? s<sub>Fe</sub> = 0.449 J/(gK)
According to the heat equation:Q = mc(T2-T1)where Q - absorbed or released heat (842.05 J), c - specific heat (0.449 J g-1 K-1), T2 - final temperature (28 oC =...
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if 842 pounds of moon samples have been collected from lunar landings, what is mass expressed in kilograms? (given 1kg =2.20lb)
383 kg
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How many grams of AgBr will form when 45.0 mL of 0.842 M CaBr2 react?
CaBr2(aq) + 2 AgNO3(aq) → 2 AgBr(s) + Ca(NO3)2(aq)
Solution:Molar mass of AgBr = 187.77 g/molmass of AgBr = M of CaBr2× V of CaBr2 / 1000ml × n of AgBr × molar mass of AgBr = 0.842molCaBr2× 45mlCaBr2/1000ml×2×187.77g/molAgBr...
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Calculate the nuclear binding energy per nucleon for the He nucleus.
(atomic masses: He = 4.0026 amu; H = 1.0078 amu; n = 1.0087 amu; 1 amu = 1.6605 x 10-27 kg)
E= mc^2.E=1.66x10-27 x (3 x 10^8)^2=1,494e-10 kJ.
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What percentage of the mass of the isotope boron-10 do its neutrons contribute? proton = 1.0073 amu neutron = 1.0087 amu electron = 5.4858×10<sup>-4</sup> amu ............%
5 neutron 5 protons and 5 electrons.neutron = 5×1.0087 = 5.0435total = 5×1.0073 + 5×1.0087 + 5×5.4858x10-4total = 5.0365 + 5.0435 + 0.0027429total = 10.0827429percentage of neutron = (neutron/total )x100=[...
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What percentage of the mass of the isotope carbon-12 do its neutrons contribute? proton = 1.0073 amu neutron = 1.0087 amu electron = 5.4858×10<sup>-4</sup> amu .............%
There are 6 protons, neutrons and electrons in the carbon-12 atom. Thus, the mass percentage of neutrons equals:6*1.0087/(6*1.0087+6*1.0073+6*5.4858*10^-4)=0.50021110822=50,021110822%.Solution: 50,02%.
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