The answer is 0.5 moles because if 1 moles of o2 is consumed to produce 2 moles of Na2O, then 0.5 moles of o2 will be consumed to produce 1 mole of Na2O.<span />
The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
Answer:
i = 2.79
Explanation:
The excersise talks about the colligative property, freezing point depression.
Formula to calculate the freezing point of a solution is:
Freezing point of pure solvent - Freezing point of solution = m . Kf . i
Let's replace data given. (i = Van't Hoff factor, numbers of ions dissolved in solution)
48.1°C - 44°C = 0.15 m . 9.78°C/m . i
4.1°C / (0.15 m . 9.78°C/m) = i
i = 2.79
In this case, numbers of ions dissolved can decrease the freezing point of a solution, which is always lower than pure solvent.
Mass % of nitrogen = mass of nitrogen*100 / total mass
= 14*100 / (1+ 14 + 32)
= 14*100 / 47
= 29.7 %
Answer:
21 g/mL
Explanation:
To solve this problem, first look at the density equation, which is D=M/V, which D stands for density, M stands for mass, and V stands for volume. When you substitute in the variables, you get D=17.5/.82, which is equivalent to 21.34. However, since we need to pay attention to the sig fig rules for multiplying, we need to have the same amount of sig figs as the value with the least amount of sig figs, which is the number .82. .82 has two sig figs, so you round down. Your answer will be 21 g/mL.
