Answer:
79.90amu
Explanation:
To calculate the average atomic mass of an bromine we must multiply the masses of its isotopes with the percent abundance divided by 100. Then you add them together for your answer.
50.69 ÷ 100 = 0.5069
49.31 ÷ 100 = 0.4931
78.9183361amu × 0.5069 = 40.00amu
80.916289amu × 0.4931 = 39.90amu
40.00amu + 39.90amu = 79.90amu
Well assuming we have all of these, earth
5.6 atm
In a container, the sum of partial pressures equals the total pressure. If the sun is 9.8 atm and the partial pressure of oxygen is 4.2 atm, the rest of the gases must account for the other 5.6 atm.
Answer:
99.56 g.
Explanation:
- The balanced reaction is:
<em>3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO₄)₂.</em>
It is clear that 3.0 moles of Ca(NO₃)₂ react with 2.0 moles of Na₃PO₄ to produce 6.0 moles of NaNO₃ and 1.0 mole of Ca₃(PO₄)₂.
- We need to calculate the no. of moles of 96.1 g of Ca(NO₃)₂:
n = mass/molar mass = (96.1 g)/(164.088 g/mol) = 0.5857 mol.
<u><em>Using cross multiplication:</em></u>
3.0 moles of Ca(NO₃)₂ produce → 6.0 moles of NaNO₃.
0.5857 moles of Ca(NO₃)₂ produce → ??? moles of NaNO₃.
∴ The no. of moles of NaNO₃ = (6.0 mol)(0.5857 mol)/(3.0 moles) = 1.171 mol.
<em>∴ The no. of grams of NaNO₃ = (no. of moles of NaNO₃)(molar mass)</em> = (1.171 mol)(84.99 g/mol) = <em>99.56 g.</em>