Answer:
A system of knowledge and the methods to find that knowledge is Science.
Answer: I dont take L´s. Only W´s so ha!!!
Explanation: Lol just messin with ya. I actually dont know what your question is but have a good one, and stay safe!! :))
Answer:
5.52atm
Explanation:
Using the pressure law formula:
P1/T1 = P2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the question, the following information were provided;
P1 = 4.72 atm
P2 = ?
T1 = -3.50°C = -3.50 + 273 = 269.5K
T2 = 42°C = 42 + 273 = 315K
Using P1/T1 = P2/T2
4.72/269.5 = P2/315
CROSS MULTIPLY
4.72 × 315 = 269.5 × P2
1,486.8 = 269.5P2
P2 = 1,486.8 ÷ 269.5
P2 = 5.52atm
Answer:
Methods of soil conservation are listed below.
Explanation:
The major sources of soil erosion include water,wind and tillage. In order to mitigate or prevent soil erosion, some of the following techniques can be implemented:
- <u>Contour Farming: </u>Planting in row patterns that run level around a hill — as opposed to the up and down the slope pattern.This reduces runoffs and consequently water erosion.
- <u>Crop Rotation:</u> This involves planting crops with high residue (e.g corn, small grains, e.t.c) in rotation,as the layer of residue would protect the topsoil.
- <u>Built in structural diversion</u> : Used often for gully control, to regulate flow of water away from the field and through designated desired paths.
- <u>Conservation Tillage</u>: This involves methods such as no-till planting, strip rotary tillage, etc, which do not allow the soil surface to be smooth and bare, but instead covered with crop residue that protects the soil from eroding forces.
The complete question is;
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.
Q1)
methyl butyrate (component of apple taste andsmell): C -58.80 % H- 9.87 %
O -31.33.%Express your answer as a chemical formula.
Q2)
vanillin (responsible for the taste and smellof vanilla): C - 63.15% H- 5.30 %
O - 31.55%Express your answer as a chemical formula.
Q1)
empirical formula is the simplest ratio of whole number of elements in the compound. as the percentages have been given, lets calculate for 100 g of compound
C H O
mass 58.80 g 9.87 g 31.33
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 58.80/12 9.87/1 31.33/16
= 4.9 =9.87 = 1.95
then divide number of moles by least number of moles - 1.95 in this case
4.9/1.95 = 2.51 9.87/1.95 = 5.06 1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
2.51x2 = 5.02 5.06x2 = 10.12 1x2 = 2
when rounded off to the nearest whole number
C - 5
H - 10
O - 2
therefore empirical formula is C₅H₁₀O₂
Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound
C H O
mass 63.15 g 5.30 g 31.55 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.15/12 5.30/1 31.55/16
=5.26 =5.30 =1.97
divide the number of moles by the least number of moles - 1.97
5.26/1.97 5.30/1.97 1.97/1.97
=2.67 = 2.69 = 1
multiply each by 3 to get numbers that can be rounded off to whole numbers
2.67x3 = 8.01 2.69x3 = 8.07 1x3 = 3
rounded off to the nearest whole numbers
C - 8
H - 8
O - 3
empirical formula = C₈H₈O₃