Answer:
C. 1 cubic foot of loose sand
Explanation:
For many objects having equal volume , surface area will be maximum
of the object which has spherical shape .
But when a sphere is broken into tiny small spheres , total surface area of all the small spheres will be more than surface area of big sphere .
Hence among the given option , surface area of loose sand will have greatest surface area . Loose sand is equivalent to small spheres .
<h2>Answer:</h2>
<h2>Explanations</h2>
The complete balanced equation for the given reaction is expressed as;
Given the following parameters
Mass of CH4 = 5.90×10^−3 g = 0.0059grams
Determine the moles of methane
According to stoichimetry, 1 mole of methane produces 2 moles of water, hence the moles of water required will be:
Determine the mass of water produced
Therefore the mass of water produced from the complete combustion of 5.90×10−3 g of methane is 1.33 * 10^-2grams
When it has been disproven
Silver chloride produced : = 46.149 g
Limiting reagent : CuCl2
Excess remains := 3.74 g
<h3>Further explanation</h3>
Reaction
silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate
Required
silver chloride produced
limiting reagent
excess remains
Solution
Balanced equation
2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)
mol AgNO3 :
= 58.5 : 169,87 g/mol
= 0.344
mol CuCl2 :
=21.7 : 134,45 g/mol
= 0.161
mol ratio : coefficient of AgNO3 : CuCl2 :
= 0.344/2 : 0.161/1
= 0.172 : 0.161
CuCl2 as a limiting reagent
mol AgCl :
= 2/1 x 0.161
= 0.322
Mass AgCl :
= 0.322 x 143,32 g/mol
= 46.149 g
mol remains(unreacted) for AgNO3 :
= 0.344-(2/1 x 0.161)
= 0.022
mass AgNO3 remains :
= 0.022 x 169,87 g/mol
= 3.74 g
Answer:
Large objects close together
