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3 years ago

30 Points worthy

Chemistry

1 answer:

OLga [1]3 years ago

5 0

Answer:

The nitrogen cycle is a cycle by which the nitrogen is converted into forms which are utilized by the living organisms and these forms circulates among the atmosphere, biosphere, and marine ecosystem. The conversion of nitrogen is through biological and physical processes.

Nitrogen Fixation is the first step in nitrogen cycle. It involves the conversion of atmospheric nitrogen into ammonia. This is done by bacteria like Azotobacter and Rhizobium.

Assimilation involves the absorption of ammonia by the plants from the soil.

Ammonification is the process by which the dead organic remains of plants and animals are decomposed by the soil bacteria to produce ammonia.

Nitrification is the process in which the ammonia is converted into nitrates by the soil bacteria Nitrosomonas and Nitrococcus. The nitrates are again converted into nitrites by the bacteria Nitrobacter.

On the basis of above description, the following is the match of steps of the nitrogen cycle.

A. Nitrification - bacteria convert nitrates into nitrogen gas.

D. Assimilation - plants and animals take in and use nitrogen compounds.

E. Denitrification - bacteria break down waste products and return nitrogen to the soil.

B. Nitrogen fixation - bacteria convert atmospheric nitrogen to ammonia.

C. Ammonification - bacteria convert ammonia to nitrites and nitrates

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How much water (H2O ) would form if 4.04 g of hydrogen (H2) reacted with 31.98 g of oxygen (O2 )?

Norma-Jean [14]

Answer:

Mass = 36 g

Explanation:

Given data:

Mass of water formed = ?

Mass of hydrogen = 4.04 g

Mass of oxygen = 31.98 g

Solution:

Chemical equation:

2H₂ + O₂ → 2H₂O

Number of moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 4.04 g/ 2 g/mol

Number of moles = 2.02 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 31.98 g/ 32 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of water with hydrogen and oxygen.

O₂ : H₂O

1 : 2

H₂ : H₂O

2 : 2

2.02 : 2.02

Number of moles of water formed by oxygen are less thus oxygen will limiting reactant.

Mass of water:

Mass = number of moles × molar mass

Mass = 2 mol × 18 g/mol

Mass = 36 g

8 0

2 years ago

Which of these best describes the difference between the formulas for nitrogen monoxide and nitrogen dioxide? Which of these bes

Pani-rosa [81]

Nitrogen monoxide has 1 oxygen atom and
Nitrogen dioxide has 2 oxygen atoms

3 0

2 years ago

How many sodium ions are in 4543.3 g of NaCl?

Stella [2.4K]

Answer:

4.68x10²⁵ ions of Na⁺

Explanation:

First of all, we dissociate the salt:

NaCl(aq) → Na⁺(aq) + Cl⁻(aq)

An aqueous solution of NaCl dissociates in chlorides anions and sodium cations. Ratio is 1:1, per 1 mol of NaCl, we have 1 mol of Na⁺

We determine the moles of salt: 4543.3 g . 1mol / 58.45 g = 77.7 moles

77.7 moles are the amount of NaCl, therefore we have 77.7 moles of Na⁺.

We count the ions:

1 mol fo Na⁺ has 6.02x10²³ ions

77.7 moles of Na⁺ must have (77.7 . 6.02x10²³) / 1 = 4.68x10²⁵ ions of Na⁺

3 0

3 years ago

Read 2 more answers

enzyme‑catalyzed, single‑substrate reaction E + S − ⇀ ↽ − ES ⟶ E + P . The model can be more readily understood when comparing t

laila [671]

Complete Question

The complete question is shown on the first uploaded image

Answer:

[S]<<KM | [S]=KM | [S]>>KM | Not true

____________ | Half of the active | Reaction rate is | Increasing

$[E_{free}]$ is about | sites are filled of | independent of | $[E_{Total}]$ will

equal to $[E_{total}]$ . | | [S] | lower KM

_____________________________________________|____________

[ES] is much | | Almost all active

lower than | | sites are filled

$[E_{free}]$ | |

Explanation:

Generally the combined enzyme $[ES]$ is mathematically represented as

$[ES] = \frac{[E_{total}][S]}{K_M + [S]}----(1)$

for Michaelis-Menten equation

Where $[S]$ is the substrate concentration and $K_M$ is the Michaelis constant

Considering the statement $[S] < < K_M$

Looking at the equation $[S]$ is denominator so it can be ignored(it is far too small compared to $K_M$ ) hence the above equation becomes

$[ES] = \frac{[E_{total}][S]}{K_M}$

Since $[S]$ is less than $K_M$ it means that $\frac{[S]}{K_M} < < 1$

so it means that $[ES] < < [E_{total}]$

What this means is that the number of combined enzymes $[ES]$ i.e the number of occupied site is very small compared to the the total sites $[E_{total}]$ i.e the total enzymes concentration which means that the free sites [ $E_{free}]$ i.e the concentration of free enzymes is almost equal to $[E_{total}]$

Considering the second statement

$[S] = K_M$

So this means that equation one would now become

$[ES] = \frac{[E_{total}][S]}{2[S]} = \frac{[E_{total}]}{2}$

So this means that half of the active sites that is the total enzyme concentration are filled with S

Considering the Third Statement

$[S] >>K_M$

In this case the $K_M$ in the denominator of equation 1 would be neglected and the equation becomes

$[ES] = \frac{[E_{total}] [S]}{[S]} = [E_{total}]$

This means that almost all the sites are occupied with substrate

The rate of this reaction is mathematically defined as

$v =\frac{V_{max}[S]}{K_M [S]}$

Where v is the rate of the reaction(also know as the velocity of the reaction at a given time t) and $V_{max}$ is he maximum velocity of the reaction

In this case also the $K_M$ at the denominator would be neglected as a result of the statement hence the equation becomes

$v = \frac{V_{max}[S]}{[S]} = V_{max}$

So it means that the reaction does not depend on the concentration of substrate [S]

For the final statement(Not True ) it would match with condition that states that increasing $[E_{total}]$ will lower $K_M$

This is because $K_M$ does not depend on enzyme concentration it is a property of a enzyme

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3 years ago

Which of the following statements about elements is not true?

guapka [62]

Answer:

Option A is not true

Explanation:

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3 years ago