Answer:
Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g.
Explanation:
Part B:
The mass of water produced when 4.86 g of butane(C4H10) react with excess oxygen is calculated as below
calculate the moles of C4H10 used = mass/molar mass
moles = 4.86g/58 g/mol =0.0838 moles
write a balanced equation for reaction
2 C4H10 + 13 O2 = 8 CO2 + 10 H2O
by use of mole ratio between C4H10 to H2O which is 2:10 the moles of
H20= 0.0838 x10/2 = 0.419 moles of H2O
mass = moles x molar mass
=0.419 molx 18 g/mol = 7.542 grams of water is formed
The atomic mass of rubidium listed in the periodic table to determine the mass of Rb−87 is 86.13 amu.
<h3>What is atomic mass?</h3>
The atomic mass is the weight of the roton neutron and electron present inside the nucleus and shells of an atom and the elements are arranged on the basis of this only.
The mass of Rb−85 is 84.9118 amu for 1 amu it will be
amu = 84.9118 / 85 = 0.99
so, the amu for Rb−87 will be,
AMU = 87 × 0.99 = 86.13 amu.
Therefore, the mass of Rb−87 is 86.13 amu for atomic mass of rubidium listed in the periodic table.
Learn more about atomic mass, here:
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Answer:
the osmolarity of the 0.85% NaCl solution is approximately equal to 300 mosmol/L.
Explanation:
Molarity= number of moles/volume of the solution in liters
volume of the solution= 100 mL
1 L= 1000 mL
100 mL= 100/1000=0.1 L
molar mass of NaCl= 58.44 g/mol
number of moles= mass in gram/gram molecular mass
= 0.85/58.44
= 0.01455
molarity= number of moles/volume of the solution in liters
= 0.01455 / 0.1
= 0.1455 M
1 M= 1000 mM
so 0.1455 M = 0.1455 × 1000
= 145.5 mM
Osmolarity is the concentration of solutes in the solution.
NaCl dissociates to Na+ and Cl-
so osmolarity of 145.5 mM NaCl= 2 × 145.5
= 291 mosmol/L
Therefore, we conclude that the osmolarity of the 0.85% NaCl solution is approximately equal to 300 mosmol/L.