28 g soil sample was heated to a constant weight of 24 g. When further heated to red hot and cooled, it weighed 18 g. What is the percentage of humus in the soil?
22.2
55.6
75.0
25.0
35.7
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Hi Guys
there is answer is not there!!, so please let the correction be known as 21.4, in this situation a smart jambites will choose the closest logical answer 22.2
24-18=6
6/28=0.2142× 100=21.42 the exact answer but the nearest to it happens to be 22.2 and obviously becomes the one to make use of
Formula =absolute value /actual value %
Absolute values are 28 and 24 while the actual value is 18.
I.E. 28-24/18 *100 =22.2
Mass of soil -28g (eq 1)
Mass of soil when heated -24g(eq 2)
Mass of soil when heated and cooled -18g(eq 3)
Substract eq 2 from eq 1.... 28-24=4
Divide answer by eq 3... 4÷18=0.222
Multply answer by 100%.... 0.222×100=22.2
pls the answer is 25
it is obvious cus the soil is wet soil and also, it is the percentage of hummus, therefore, we get the hummus lost which is 24-8=6 and then the total mass of the soil 24 multiplied by 100
mass of wet soil = 28g
mass of dry soil = 24g
mass of red hot soil = 18g
Humus is burnt away when dried soil is heated to red hot soil.
Therefore
mass of humus = 24-18 =6g
Percentage of humus = (mass of humus/mass of dried soil) * 100/1
(6/24)*100/1
=25%
ANS = D
mass of crucible = M1. . mass of crucible + fresh soil= M2. mass of crucible + dry soil before strong heating = M3. mass of crucible + dry soil after strong heating = M4. m1 = 0 m2 = 28 m3 = 24 m4 =18. percentage of humus in the soil is m3 - m4/ m2 -m1 *100 24 -18/28 -0 * 100 = 6/28 *100 = 21.4
To determine the percentage of humus in the soil, follow these steps:
1. Initial mass of soil sample = 28 g
2. Mass after heating to constant weight (removal of water) = 24 g
This means the mass of water lost = g
3. Mass after further heating to red hot and cooling = 18 g
This means the mass of organic matter (humus) lost = g
4. Percentage of humus is calculated as:
\frac{\text{Mass of humus lost}}{\text{Initial mass of soil sample}} \times 100
\frac{6}{28} \times 100 = 21.43\%
The closest option is 22.2% (A).
@benjamin how can ur answer be 22.2 when ur solving equals 21.4....... mind u....... 6/28=0.214×100=21.4
The answer is 22.2
first find the difference between 28 & 24
(28-24=4)
4/18×100=22.2222222....
mass of crucible = M1. . mass of crucible + fresh soil= M2. mass of crucible + dry soil before strong heating = M3. mass of crucible + dry soil after strong heating = M4. m1 = 0 m2 = 28 m3 = 24 m4 =18. percentage of humus in the soil is m3 - m4/ m2 -m1 *100 24 -18/28 -0 * 100 = 6/28 *100 = 21.4 modern biology page 114
