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2 years ago

Why is ammonification an important process?

Chemistry

1 answer:

Nookie1986 [14]2 years ago

8 0

Answer: B) Fungi convert nitrogen in dead plants into ammonium, which converts proteins into ammonia.

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The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

8 0

2 years ago

Estimate the heat of neutralization of H3PO4 in kJ/mol given these data: Heat transferred = 2.2 x 102 J Moles of H3PO4 = 1.5 x 1

barxatty [35]

<span>write out the balance equation
3NaOh+H3PO4->Na3PO4+3H2O
You are given everything needed to calculate
q=heat transfer=2.2*10^2,
H3PO4 moles= 1.5*10^-3,
NaOH moles=5.0*10^-3
equation is deltaHneutraliztion=q/Moles of limiting reagent H3PO4 is limiting reagent because lowest moles, and is used up first
Now plug in variables
DeltaH=2.2*10^2(1.5*10^3)= 146.67kj/mole
Notice we had to convert J to kj,</span>

4 0

2 years ago

Read 2 more answers

Perform the following operation and express the answer in scientific notation. 7.0x10^3x1.1x10^7

dimulka [17.4K]

Answer:

The answer is: $7.7\times10^{10}$

Explanation:

Scientific notation is of the kind: $a.bc\times10^{x}$

The given calculation:

$7.0\times10^{3}\times1.1\times10^{7}=(7\times1.1)\times10^{(3+7)}$

$=7.7\times10^{10}$

Therefore, the answer in scientific notation is: $7.7\times10^{10}$

4 0

2 years ago

A cracker crushed in water will test positive with iodine but negative with benedict’s solution. A

german

A cracker which contains starch,is test positive with iodine solution but not Benedict's solution. This is because Benedict's solution is used to test for reducing sugars like glucose,galactose,fructose,maltose and lactose.In this case, the cracker is added with amylase enzyme which hydolyses starch into maltose,thus benedict's solution is test positive

7 0

2 years ago

(a) The student dissolves the entire impure sample of CuSO4(s) in enough distilled water to make 100.mL of solution. Then the st

Talja [164]

Answer:

Explanation:

Given parameters :

Volume of solution = 100mL

Absorbance of solution = 0.30

Unknown:

Concentration of CuSO₄ in the solution = ?

Solution:

There is relationship between the absorbance and concentration of a solution. They are directly proportional to one another.

A graph of absorbance against concentration gives a value of 0.15M at an absorbance of 0.30.

The concentration is 0.15M

Also, we can use: Beer-Lambert's law;

A = ε mC l

where εm is the molar extinction coefficient

C is the concentration

l is the path length

Since the εm is not given and assuming path length is 1;

Then we solve for the concentration.

3 0

3 years ago