All categories

3 years ago

3. What is found around the roots of some plants that turns nitrogen from the form we cannot use to the form we can use?

Chemistry

1 answer:

dsp733 years ago

5 0

Answer:

Bacteria

Explanation:

Some plants, specifically leguminous plants form a symbiotic relationship with certain bacteria species called NITROGEN-FIXING BACTERIA in their root nodules. In this relationship, the bacteria species e.g nitrosomonas, nitrobacter etc help the plants fix atmospheric nitrogen (N2) or ammonium compounds, which are forms of nitrogen that are unusable to the plant into nitrates, a usable form.

Therefore, BACTERIA are the organisms found around the roots of these set of plants (legumes) that turns nitrogen from the form we cannot use to the form we can use.

You might be interested in

See the picture to answer the question

amid [387]

Above is a potential energy curve of a reaction. It depicts conversion of reactant to product via transition state.

When a catalyst is added to the reaction system, energy barrier of reaction decreases.

It must be noted that energy barrier reaction is defined as energy difference between reactant and transition state.

In present case, energy of reactant is 200 kj, while that of transition state (in absence of catalyst) is 650 kj

Thus, energy barrier of reaction is 650 - 200 = 450 kj

<span>Hence, system must absorb 450 kj of energy for the reaction to start, if no catalyst was used</span>

4 0

3 years ago

An electrode has a potential of 1.201 V with respect to a saturated silver-silver chloride electrode. What would the electrodes

fredd [130]

Answer:

The potential wrt. calomel is 1.254 V

Explanation:

Given:

Potential wrt. silver chloride $E_{Ag} = 1.201$ V

Potential wrt. saturated silver chloride $E = 0.197$ V

Potential wrt. SCE $E_{Hg} = 0.241$ V

Now potential wrt. hydrogen is given by,

$= 1.201- 0.197$

$= 1.004$ V

And we find for potential wrt. calomel,

$=$ potential wrt. hydrogen + potential wrt. SEC

$= 1.004 +0.241$

$= 1.254$ V

Therefore, the potential wrt. calomel is 1.254 V

7 0

3 years ago

Which member of each pair is the stronger base? part a ethylamine or aniline ethylamine or aniline ethylamine aniline request an

Vladimir79 [104]

Lowery-Bronsted theory is applied here. Acc. to this theory Base accepts protons and Acids donate proton.

Part 1:
Aniline is less basic than ethylamine because the lone pair on nitrogen (which accepts proton) is not localized. It resonates throughout the conjugated system of phenyl ring. Hence due to unavailability of electrons for accepting proton it is less basic compare to ethylamine. In ethyl amine the lone pair of electron is localized and available to abstract proton.

Part 2:
In this case the alkyl groups attached to -NH₂ (in ethylamine) and -O⁻ (in ethoxide are same (i.e. CH₃-CH₂-). Ethoxide is more basic than ethylamine because ethoxide is a conjugate base of ethanol (pKa value of ethanol = 15.9 very weak acid) and the conjugate base of weak acid is always a strong base. Secondly, the oxygen atom more Electronegative than Nitrogen atom can attract more electron cloud from alkyl group as compared to Nitrogen in ethylamine. Hence, oxygen in ethoxide attains greater electron cloud than the nitrogen in ethylamine. Therefore, it is more basic than ethylamine.

5 0

3 years ago

Suppose you want to find out which dog food your dog likes best. Write about the experiment you would design. What variables do

denis-greek [22]

By giving the dog 2 or more different dog foods and seeing what one he eats the must of and the variables u control is how much you give the dog and the hypothesis is if I give the dog two or more different kinds of dog food than we will see what one he/she like the most

3 0

3 years ago

Plot your values of ln(Ksp) vs. 1/T and find the slope and y-intercept of the best fit line. Use the equation for the best fit l

labwork [276]

Answer:

a) The slope of the line of best fit plot = -12629.507

b) ΔH∘ = 105 kJ

c) Intercept of the line of best fit plot = 39.099

d) ΔS∘ = 325.1 J/K

e) Option A is correct.

Solubility will increase as temperature increases, because as T increases the (−ΔH∘/RT) term becomes smaller therefore K will get larger.

f) Option D is correct. All of the options are correct.

Explanation:

The complete question is presented in the first attached image to this question. This complete question has the data readings required to plot the graph.

The second attached image has the plotted graph and the regression analysis to obtain the line of best fit.

The equation of the line of best fit obtained is

y = -12629.507x + 39.099

Comparing the given expression for the question with the equation of a straight line

ln (K) = (−ΔH∘/RT) + (ΔS∘/R)

y = mx + c

y = In K

Slope = m = (−ΔH∘/R)

x = (1/T)

Intercept = c = (ΔS∘/R)

So, to answer the question now

a) The slope of the line of best fit plot = -12629.507

b) Slope = (−ΔH∘/R)

(−ΔH∘/R) = -12629.507

But R = molar gas constant = 8.314 J/mol.K

ΔH∘ = 12629.507 × 8.314 = 105,001.721198 J = 105,002 J = 105 kJ

c) Intercept of the line of best fit plot = 39.099

d) Intercept = (ΔS∘/R)

(ΔS∘/R) = 39.099

ΔS∘ = 39.099 × 8.314 = 325.069086 J/K = 325.1 J/K

e) Do you expect the solubility of Borax to increase or decrease as temperature increases?

Solubility will increase as temperature increases, because as T increases the (−ΔH∘/RT) term becomes smaller therefore K will get larger.

f) Why was it necessary to make sure that some solid was present in the main solution before taking the samples to measure Ksp? Select the option that best explains why.

A. To make sure no more sodium borate would dissolve in solution.

B. To ensure the dissolution process was at equilibrium.

C. To make sure the solution was saturated with sodium and borate ions.

D. All of the above

Hope this Helps!!!

5 0

3 years ago