Answer:
225g
Explanation:
250mL x 0.9g/mL = 225g
Answer:
225g
Explanation:
d=m/ml
m= d.ml
m=0,9g/ml.250ml
m=225g
Molecular complexes in the cristae contain electron transport carriers that move _________ into the intermembrane space. The first electron transport carrier accepts an ________ from _________ . A second carrier accepts an ___________ from _________ . After the ________have travelled through the chain they are finally accepted by ________ and 2 __________ to produce water. Going from high to low concentration, _________ flow through ATP synthase like a water wheel to produce ~32 ________ molecules.
Answer:
a.) Hydrogen
b.) electron
c.) NADH⁺ + H⁺
d.) electron
e.) FADH₂
f.) electron
g.) hydrogen ions
h.) oxygen
i.) hydrogen ions
j.) ATP
Explanation:
Molecular complexes in the cristae contain electron transport carriers that move Hydrogen into the intermembrane space.
The first electron transport carrier accepts an electron from NADH⁺ + H⁺
A second carrier accepts an electron from FADH₂ .
After the electron have travelled through the chain they are finally accepted by hydrogen ions and oxygen to produce water.
Going from high to low concentration, hydrogen ions flow through ATP synthase like a water wheel to produce ~32 ATP molecules.
What is the molarity of a solution that contains 152 g NaCl in 4.00 L solution?
Answer:
200.0lg
Explanation:
please give a brainliest
Civic participation is best described as the act of
avoiding harming other community members.
meeting one's legal responsibilities.
voting in every type of election.
going above and beyond one's legal duties.
Which is one of Edwin Hubble’s findings that supports the big bang theory?
Answer:
meeting ones legal responsibility
Zack has noticed a lot of changes in his body. He has gained muscle mass and started to grow facial hair. His voice is also getting deeper.
In which stage of development is he?
childhood
adolescence
early adulthood
late adulthood
Answer:
b
Explanation:
The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the same empirical formula as propene, C3H6
Sample Mass of carbon Mass of hydrogen
A 60g 12g
B 72g 12g
C 84g 10g
D 90g 10g
a. Sample A
b. Sample B
c. Sample C
d. Sample D
Answer:
Sample B
Explanation:
In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.
Let's do Sample A:
C: 60 g; H: 12 g
1. Calculate moles:
We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):
C: 60 / 12 = 5
H: 12 / 1 = 12
2. Determine number of atoms in the formula
In this case, we just divide the lowest moles obtained in the previous part, by all the moles:
C: 5 / 5 = 1
H: 12 / 5 = 2.4 or rounded to two
3. Write the empirical formula:
Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:
C₁H₂ = CH₂
Therefore, sample A is not the same as propene.
Sample B:
C: 72 g H: 12 g
Following the same steps, let's determine the empirical formula for this sample
C: 72 / 12 = 6 ---> 6 / 6 = 1
H: 12 / 1 = 12 ----> 12 / 6 = 2
EF: CH₂
Sample C:
C: 84 g H: 10 g
C: 84 / 12 = 7 ----> 7 / 7 = 1
H: 10 / 1 = 10 ----> 10 / 7 = 1.4 or just 1
EF: CH
Sample D
C: 90 g H: 10 g
C: 90 / 12 = 7.5 -----> 7.5 / 7.5 = 1
H: 10 / 1 = 10 -------> 10 / 7.5 = 1.33 or just 1
EF: CH
Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:
C3H6 -----> CH₂
Therefore, sample B is the one that match completely. Sample B would be the one.
Hope this helps
The celestial bodies in space move about in predictable paths called blank . As a result, a planet is able to blank around the Sun. Other objects in the solar system also orbit the Sun. These objects include blank and comets.
Answer:
Gravity
Explanation:
All matter exerts a force, which is called , that pulls all other matter towards its center. The strength of the force depends on the mass of the object: the Sun has more gravity than Earth, which in turn has more gravity than an apple. Also, the force weakens with distance. Objects far from the Sun won’t be influenced by its gravity.
The exhaust from car engines spreads throughout our atmosphere and adds to pollution. What is this an example of?
A.
An increase in enthalpy
B.
An increase in thrust
C.
An increase in thermodynamics
D.
An increase in entropy
(D is the correct answer!)
Answer: When exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
Explanation:
A degree of randomness in the molecules of a substance is called entropy.
When exhaust from car engines spreads throughout our atmosphere then it means the molecules are moving at a faster speed due to which they spread into the atmosphere rapidly.
This means that the entropy of exhaust is increasing.
Thus, we can conclude that when exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.
An increase in entropy
In each pair of compounds, pick the one with the higher vapor pressure at a given temperature. Explain your reasoning.
a. CH4 or CH3CI
b. CH3CH2CH2OH or CH3OH
c. CH3OH or H2CO
Answer:
The answer is:
[tex]PART \ A: CH_4\\\\PART \ B: CH_30H\\\\PART \ C: H_2CO[/tex]
Explanation:
Parts A:
The vapor pressure is higher in [tex]CH_4[/tex] because it is non-polar, while [tex]CH_3Cl[/tex] is polar. [tex]CH_4[/tex] has a lower molar weight as well.
Part B:
Although hydrogen bonding is found in both commodities, the vapor pressure is higher because of the smaller molar mass of [tex]CH_30H[/tex].
Part C:
[tex]H_2CO[/tex] does not show hydrogen due to the increased vapor pressure
[tex]CH_3OH[/tex] bonding.
Calculate the mass of sucrose needed to prepare a 2000 grams of 2.5% sucrose solution.
Answer:
50 g Sucrose
Explanation:
Step 1: Given data
Mass of solution: 2000 gConcentration of the solution: 2.5%Step 2: Calculate the mass of sucrose needed to prepare the solution
The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:
2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose
A solution is prepared by dissolving 16.90 g of ordinary sugar (sucrose, C12H22O11, 342.3 g/mol) in 40.90 g of water. Calculate the boiling point of the solution. Sucrose is a nonvolatile nonelectrolyte.
Answer:
Explanation:
The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT
ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution
Kb for water = .51°C /m
moles of sugar = 16.90 / 342.3
= .04937 moles
m = moles of sugar / kg of water
= .04937 / .04090
= 1.207
ΔT = Kb x m
= .51 x 1.207
= .62°C .
So , boiling point of water = 100.62°C .
Sample A: 300 mL of 1M sodium chloride
Sample B: 145 mL of 1.5 M sodium chloride
Which sample contains the larger concentration of sodium chloride?
Answer:
Sample A
Explanation:
Molarity of a solution can be calculated using the formula:
Molarity = number of moles (mol) ÷ volume (vol)
For Sample A:
V = 300ml = 300/1000 = 0.3 L
Molarity = 1M
n = number of moles (mol)
1 = n/0.3
n = 0.3moles
For Sample B:
V = 145 mL = 145/1000 = 0.145L
Molarity = 1.5 M
n = number of moles
1.5 = n/0.145
n = 1.5 × 0.145
n = 0.22 moles
Based on the above results (moles), sample A with 0.3 moles contains the larger concentration of sodium chloride.
One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Over-reliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the C12 and C 13 isotopes in samples of humans remains. Suppose you use a velocity selector to obtain single ionized (missing one electron) atoms of speed 8.50 km/s and what to bend them within a uniform magnetic field in a semicircle of diameter 25.0 cm for the C12 . The measured masses of these isotopes are 1.99 × 10^-26 kg ( C12 ) and 2.16 × 10^-26 kg ( C13).
Required:
a. What strength of magnetic field is required? (b) What is the diameter of the 13C semicircle?
b. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?
Answer:
a. i. 8.447 × 10⁻³ T ii. 27.14 cm
b. i. 2.14 cm ii. It is easily detectable.
Explanation:
a.
i. What strength of magnetic field is required?
Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have
F = F'
Bqv = mv²/r
B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m
So,
B = mv/rq
B = 1.99 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (12.5 × 10⁻² m × 1.602 × 10⁻¹⁹ C)
B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)
B = 0.8447 × 10⁻² kg/sC)
B = 8.447 × 10⁻³ T
(ii) What is the diameter of the 13C semicircle?
Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have
F = F'
Bq'v = mv²/r'
r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and = d/2 = 12.5 cm = 12.5 × 10⁻² m
So, r' = mv/Be
r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)
r' = 18.36 × 10⁻²³ kgm/s ÷ 13.5321 × 10⁻²² TC)
r' = 1.357 × 10⁻¹ kgm/TC)
r' = 0.1357 m
r' = 13.57 cm
Since diameter d' = 2r', d' = 2(13.57 cm) = 27.14 cm
b.
i. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle?
Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm
ii. Is this distance large enough to be easily observed?
This distance of 2.14 cm easily detectable since it is in the centimeter range.
30. Which of the following atoms would form
a trivalent anion?
A. Aluminium
B. Oxygen
C. Lithium
D. Nitrogen
Aluminium and Nitrogen!
Question 6
4 pts
6) Chromium crystallizes in a body-centered cubic unit cell. If the length of an edge of the unit cell
is 289 pm, what is the density (in g/cm3) of chromium?
Show your work on a paper and upload it here.
Upload
Choose a File
Answer:
7.15g/cm³
Explanation:
To solve this question we must know that a body-centered cubic unit cell contains 2 atoms.
The volume of the unit cell is:
289pm = (289x10⁻¹²m)³ =
2.414x10⁻²⁹m³ * (1cm³ / 1x10⁻⁶m³) = 2.414x10⁻²³cm³
And the mass is -Molar mass Chromium = 51.9961g/mol:
2atoms * (1mol / 6.022x10²³atoms) * (51.9961g / mol) =
1.727x10⁻²²g
The density is:
1.727x10⁻²²g / 2.414x10⁻²³cm³ =
7.15g/cm³Boyle's Law represents the relationship of volume as pressure is changed. The constants of Boyle's Law are
temperature and number of molecules of gas
Answer:
True.
Explanation:
Hello!
In this case, according to the ideal gas equation:
[tex]PV=nRT[/tex]
It turns out possible for us to obtain the Boyle's law, as a directly proportional relationship between pressure and volume, considering that both moles and temperature remain constant; in such a way, we write the following:
[tex]PV=k[/tex]
Therefore, for an initial (1) and final state (2) we are able to write:
[tex]P_2V_2=P_1V_1[/tex]
Regards!
a change in an objects position is called
Answer: are there any answer choices? If not I think MOTION
Explanation: hope that’s right, have a great day!! :)
A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the temperature changed to:
(a) -5 ºC
(b) 95 ºF
(c) 1095 K
Answer: The new volume at different given temperatures are as follows.
(a) 109.81 mL
(b) 768.65 mL
(c) 18052.38 mL
Explanation:
Given: [tex]V_{1}[/tex] = 571 mL, [tex]T_{1} = 26^{o}C[/tex]
(a) [tex]T_{2} = 5^{o}C[/tex]
The new volume is calculated as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL[/tex]
(b) [tex]T_{2} = 95^{o}F[/tex]
Convert degree Fahrenheit into degree Cesius as follows.
[tex](1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C[/tex]
The new volume is calculated as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL[/tex]
(c) [tex]T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C[/tex]
The new volume is calculated as follows.
[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL[/tex]
How many sigma bond and pi bond are present
When water is added to a mixture Na_2O_2(s) and S(s), a redox reaction occurs, as represented by the equation below.
2NaO_2(s) + 2 H_2O(l) rightarrow 4 NaOH (aq) + SO_2(aq) Delta H_298 degree = -610 kJ/mol_rxn; Delta S_298 degree = -73 J/(K mol_rxn)
Which of the following statements about the thermodynamic favorability of the reaction at 298 K is correct?
(A) It is thermodynamically unfavorable
(B) It is thermodynamically favorable and is driven by Delta S degree only.
(C) It is favorable and is driven by Delta H degree only.
(D) It is thermodynamically favorable and is driven by both Delta H degree and Delta S degree
Answer:
(C) It is favorable and is driven by ΔH° only.
Explanation:
Let's consider the following balanced equation.
2 Na₂O₂ + S + 2 H₂O → 4 NaOH + SO₂
To determine whether it will be favorable or not at 298 K, we need to calculate the standard Gibbs free energy (ΔG°).
If ΔG° < 0, the reaction will be favorable.If ΔG° > 0, the reaction will be unfavorable.We can calculate ΔG° using the following expression.
ΔG° = ΔH° - T.ΔS°
As we can see from the expression above, the favorability will be driven if ΔH° < 0 and if ΔS° < 0. Since ΔH° = -610 kJ/mol and ΔS° = -73 J/K.mol, the favorability will be driven by ΔH°. Now, let's calculate the overall favorability.
ΔG° = ΔH° - T.ΔS°
ΔG° = -610 kJ/mol - 298 K.(-0.073 kJ/K.mol) = -588 kJ/mol
The reaction is favorable and is driven by ΔH° only.
Solid lead acetate is slowly added to 75.0 mL of a 0.0492 M sodium sulfate solution. What is the concentration of lead ion required to just initiate precipitation?
Answer:
The concentration of lead ion required to just initiate precipitation is -[tex]2.37\times10^-^5 M[/tex]
Explanation:
Lets calculate -:
Solubility equilibrium -: [tex]PbI_2(s)[/tex] ⇄ [tex]Pb^2^+ (aq) + 2I^- (aq)[/tex]
Solubility product of [tex]PbI_2[/tex] ,[tex]Q=[Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][I^-]^2_i_n_i_t_i_a_l[/tex] [tex]=9.8\times10^-^9[/tex]
Concentration of [tex]I^-[/tex][tex]=[KI]=0.0492M[/tex]
When the ionic product exceeds the solubility product , precipitation of salt takes place .
[tex]Q_s_p\geq K_s_p[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][I^-]^2_i_n_i_t_i_a_l[/tex] [tex]\geq 9.8\times10^-^9[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex][0.0492]^2[/tex] [tex]\geq 9.8\times10^-^9[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq \frac{9.8\times10^-^9}{[0.0492]^2}[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq[/tex] [tex]\frac{9.8\times10^-^9}{2.42\times10^-^3}[/tex]
[tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq[/tex] [tex]2.37\times10^-^5 M[/tex]
Thus , [tex]PbI_2[/tex] will start precipitating when [tex][Pb^2^+]_i_n_i_t_i_a_l[/tex] [tex]\geq 2.37\times10^-^5 M[/tex].
What type of a liquid will have a pH value equal to 12?
Basic
Neutral
Strong acid
Weak acid
Answer:
weak acid
Explanation:
Answer: Basic is correct!!!!!
Explanation:
"At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature." This is -
Answer: B
Explanation:
According to Charle's law, at constant pressure the volume of a fixed mass of a ga is directly proportional to its absolute temperature.
At constant pressure, V∝T.
a sample of mg(hco3)2 contains 1.8 moles of oxygen atom find the number of carbon atoms in the given sample
please solve fast very much urgent
Answer:
0.60 moles of atoms of carbon
Explanation:
Step 1: Given data
Chemical formula of the compound: Mg(HCO₃)₂Moles of oxygen atoms: 1.8 molesStep 2: Calculate the number of carbon atoms in the given sample
According to the chemical formula of the compound, the molar ratio of C to O is 2:6, that is, there are 2 moles of atoms of C every 6 moles of atoms of O. The number of moles of atoms of C is:
1.8 mol O × 2 mol C / 6 mol O = 0.60 mol C
A percent is a ratio expressed per 100 units. If you have 20 balls, and 1 is blue, you can say that 5% (or 5 out of 100 balls) is blue. Similarly, if a solution is 5% KI by mass, you can say that 5 g per 100 g solution is KI. When mass percent is used as a conversion factor, it can be expressed in the units required to solve the problem provided that the ratio remains constant. If the numerator is divided by 100, the denominator must also be divided by 100. Two examples are: express the grams of solute per 100 grams of solution, or express the centigrams (cg) of solute per 1 g of solution The latter example works because 100 cg is equivalent to 1 g. Complete the following statements regarding how many grams of KI are found in either 100 g or 1 kg of a 2.5% solution of KI.
a. A 2.5% (by mass) solution concentration signifies that there is ______ of solute in every 100 g
b. Therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be ________
c. A solution mass of 1 kg is _______ times greater than 100 g, thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of
Answer:
a. 2.5 g
b. 0.025
c. i. 10 times ii 25 g of KI
Explanation:
a. A 2.5% (by mass) solution concentration signifies that there is ______ of solute in every 100 g
Since % by mass = mass of solute, m/mass of solvent,M × 100 %
2.5 % = m/M × 100%
since M = 100 g,
2.5 % = m/100 × 100 %
m/100 = 2.5/100
m = 2.5 g
b. Therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be ________
The mass ratio = mass of solute/mass of solvent
= m/M
= 2.5 g/100 g
= 0.025
c A solution mass of 1 kg is _______ times greater than 100 g, thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of
i. A solution mass of 1 kg is _______ times greater than 100 g
Since 1 kg = 1000 g, 1 kg/100 g = 1000 g/100 g = 10
So,1 kg is 10 times greater than 100 g.
ii. Thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of KI
So, if M = 1 kg = 1000 g, we find m
Since % by mass = mass of solute, m/mass of solvent,M × 100 %
2.5 % = m/M × 100%
since M = 1000 g,
2.5 % = m/1000 × 100 %
m/1000 = 2.5/100
m = 2.5/100 × 1000 g
m = 2.5 × 10 g
m = 25 g
1.the volume of air in a mountain bicycle tire is constant. the initial pressure and temperature are 3.14 atm and 26 degrees celsius.
a) as the bike is ridden friction with the road causes the tire to heat up. at what kelvin temperature will the gas in the tire have a pressure of 3.96 atm
b) explain why your answer makes sense at the particle level. use the words collisions, force, and frequency in your responde.
Answer:
Explanation:
Givens
P1 = 3.14 atm
T1 = 26 + 273 = 299
P2 = 3.96
T2 = ?
Equation
P1 / T1 = P2 / T2
Solution
3.14 / 299 = 3.96 / x Cross multiply
3.14 x = 299 * 3.96 Divide by 3.14
x = 299*3.96 / 3.14
x = 1184.04/3.14
x = 377.08 oK Take off 273 to get the degrees Centigrade.
x = 104.08 oC
Part B
If you want the temperature to increase, the relationship is a direct one. That means as the temperature goes up, the pressure will as well. So for a bicycle, it makes sense that the volume remains constant. That makes the formula above without volume because the volume would cancel out on both sides of the equation.
So as the pressure increases, the particles making up the air begin to respond by the molecules moving faster. Temperature really is the a measurement of how fast molecules are moving. So the momentum is increasing. When the particles hit something, it is with a much greater force and that increases the apparent pressure.
When liquid water freezes into solid ice in the freezer,
Answer:
What about it? I don't get the question
4. Eggs are present in the ovaries when people are born, and after they go
through puberty. *
OTrue
OFalse
Calculate the mass of sucrose needed to prepare a 2000 grams of 2.5% sucrose solution.
Answer:
mass of sucrose = 17.115 grams
Explanation:
Given that:
mass = 2000 grams
molar mass of sucrose = 342.3 g/mol
2.5% of the sucrose soltion.
Let assume we are given 1 M of the sucrose solution;
2.5% of 1 M = 0.025 M
∴
Molarity = ( mass/molar mass )(1000/V)
mass = (0.025 * 342.3 * 2000)/(1000)
mass of sucrose = 17.115 grams
A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next few minutes?
-Molecules in both the metal and the surrounding air will start moving at lower speeds. -Molecules in both the metal and the surrounding air will start moving at higher speeds. -The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up.
-The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Answer:
until the next harvest, and seed must be held for the next season's ... successful grain storage is the moisture content of the crop. ... or both. If ambient temperatures are low, then air alone may cool the ... allow some of the drying to take place naturally while the crop ... employed to cool grain that has been placed in storage.
What is the usable form of nitrogen that is found in the ground?