Answer:
The six-row combin harvests 285 bushels of corn per hour.
Explanation:
The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three.
If the relationship between the magnitudes is direct, you can apply the following formula:
a ⇒ b
c ⇒ x
So: 
In this case you can apply it in the following way: if 56 pounds is equal to 1 bushel, 266 pounds to how many bushels is it?
x=4.75
This is, a six row combin harvests 4.75 bushels of corn per min.
So, being 1 hour equal to 60 minutes, you can apply the following rule of three: if in 1 minute a six-row harvester harvests , in 60 minutes how many bushels of corn it harvests?
bushels of corn= 285
<u><em>
The six-row combin harvests 285 bushels of corn per hour.</em></u>
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
What are the answer choices?
1.54×10 −10
one and fifty four-hundreths times ten to the power of negitiive 10
Answer:
If we assume that there will be enough Hydrogen for the reaction to occur, then there will be 8 moles of NH
Explanation:
The balanced equation will look like this:
4N2 + 4H2 -> 8NH
