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3 years ago

Write the chemical formula for a molecule of non cyclic amp.

Chemistry

1 answer:

RSB [31]3 years ago

6 0

The structural formula shown in the picture is the cyclic AMP or cyclic adenosine monophosphate. It is an essential component in RNA synthesis and metabolic processes. When it becomes non-cyclic after enzymatic reaction, you would have to disconnect the cyclic rings. When you do, you add 2 H atoms, one for each terminal end. The formula for cyclic AMP is C₁₀H₁₄N₅O₇P. Since there are three cyclic rings, you would have to add 6 hydrogen atoms. Thus, the formula for noncylic AMP is C₁₀H₂₀N₅O₇P.

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A voltaic cell is constructed from an Ni2+(aq)−Ni(s)Ni2+(aq)−Ni(s) half-cell and an Ag+(aq)−Ag(s)Ag+(aq)−Ag(s) half-cell. The in

eduard

Answer:

+1.03 V

Explanation:

The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).

The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:

Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V

Ag⁺(aq) + e⁻ → Ag(s) E° = +0.80 V

The value is calculated by a spontaneous reaction, in which the cell with the greater E° is reduced (gain electrons), and the other is oxidized (loses electrons). So, Ag⁺ reduces.

emf = E°reduces - E°oxides

emf = 0.80 - (-0.23)

emf = +1.03 V

8 0

3 years ago

An 8 oz. bottle of energy drink contains 6.0 g of protein, 2.0 g of fat, and 16.3 g of carbohydrate. The fuel value of this ener

valina [46]

Answer: The correct option is d) 460 kJ

Explanation:

We are given:

Content of fat in energy drink = 2.0 g

Content of protein in energy drink = 6.0 g

Content of carbohydrate in energy drink = 16.3 g

Also,

The fuel value of fat = 38 kJ/g

The fuel value of protein = 17 kJ/g

The fuel value of carbohydrate = 17 kJ/g

So, the fuel value of the energy drink will be:

Total fuel value = $(2.0g\times 38 kJ/g)+(6.0g\times 17 kJ/g)+(16.3g\times 17 kJ/g)$

Total fuel value = $[76+102+277]=460kJ$

Hence, the correct option is d) 460 kJ

4 0

3 years ago

Explain why the containers are heavy

vfiekz [6]

Answer:

there is no picture so we can't tell why the containers are heavy

7 0

2 years ago

Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

6 0

3 years ago

When this element was discovered, it exhibited luster and malleability, and it reacted very vigorously with water. This element

TEA [102]

Answer:

a) alkali metals

Explanation:

The element described above definitely belonged to the alkali metals, the first group on the periodic table. They show the properties indicated in the text.

They are metals because only metals are lustrous and malleable. This eliminates the possibility of them being halogens and noble gases.
Only group 1 metals reacts vigorously with water to form alkali.
Alkali are aqueous solutions that are basic in nature.
The reactivity of group 1 metals is due to their one extra electrons in the outer most shell.
These electrons are easily and readily lost in order for such atoms to gain stability and replicate the nearest noble gases.
The most reactive metal belongs to this group elements.
This is why it is nearly impossible to find them occurring alone in free state.
Some of the elements in this group are Li, Na, K, Rb, Cs and Fr.
Transition metals have variable oxidation states and some can be found alone in nature.

8 0

2 years ago