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Evgen [1.6K]
3 years ago
12

Please help me immediately I will mark you as brainiest.

Chemistry
1 answer:
IrinaVladis [17]3 years ago
5 0
There's 18 atoms in ammonium phosphate
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Which of the following is true about the offspring of an organism X that reproduces asexually?
Mekhanik [1.2K]
It would be that "the offspring would have identical DNA to X. When an organism reproduces asexually, it creates an exact copy of its genetic material to pass on to its offspring.


Hope this helped :)
8 0
3 years ago
Read 2 more answers
A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,
Dafna1 [17]

Answer : The concentration of A,B\text{ and }C at equilibrium are 0.132 M, 0.232 M  and 0.168 M  respectively.

Explanation :

The given chemical reaction is,

A(aq)+B(aq)\rightleftharpoons C(aq)

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

\Delta G^o=-RT \ln k

where,

\Delta G^o = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above relation, we get:

-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k

k=5.45

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

                          A(aq)+B(aq)\rightleftharpoons C(aq)

Initially               0.30      0.40         0  

At equilibrium  (0.30-x) (0.40-x)     x

The expression of equilibrium constant will be,

k=\frac{[C]}{[A][B]}

5.45=\frac{x}{(0.30-x)\times (0.40-x)}

By solving the term x, we get

x=0.168\text{ and }0.716

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of A at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of B at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of C at equilibrium = x = 0.168 M

3 0
4 years ago
The atomic number of nitrogen is 7. How many protons, neutrons, and electrons make up an atom of Nitrogen-15?
earnstyle [38]
<h2><u>Answer:</u></h2>

A nonpartisan iota of Nitrogen has a mass of 18. There are 7 protons in the core of this iota. What number of neutrons, complete electrons, and valence electrons are available  

Nitrogen 15 has a nuclear mass of 15. The mass number is # protons in addition to # of neutrons, so for N-15 mass is 15 and the protons are dependably 7 so there must be 15-7=8 neutrons. N-15 has 7 electrons since it has 7 protons and p = e.

6 0
3 years ago
Ethers are almost always used as solvents for Grignard reactions, all of the reasons why they work so well are not fully underst
vfiekz [6]

Answer:

Ether is used as a solvent because it is aprotic and can solvate the magnesium ion.

Explanation:

Solubility in Water

Because ethers are polar, they are more soluble in water than alkanes of a similar molecular weight. The slight solubility of ethers in water results from hydrogen bonds between the hydrogen atoms of water molecules and the lone pair electrons of the oxygen atom of ether molecules.

Ethers as Solvents

Ethers such as diethyl ether dissolve a wide range of polar and nonpolar organic compounds. Nonpolar compounds are generally more soluble in diethyl ether than alcohols because ethers do not have a hydrogen bonding network that must be broken up to dissolve the solute. Because diethyl ether has a moderate dipole moment, polar substances dissolve readily in it.

Ethers are aprotic. Thus, basic substances, such as Grignard reagents, can be prepared in diethyl ether or tetrahydrofuran. These ethers solvate the magnesium ion, which is coordinated to the lone pair electrons of diethyl ether or THF. Figure attached, shows the solvation of a Grignard reagent with dietheyl ether.

The lone pair electrons of an ether also stabilize electron deficient species such as BF3 and borane (BH3). For example, the borane-THF complex is used in the hydroboration of alkenes (Section 1

5 0
4 years ago
A 0.200 M solution of a weak acid, HA, is 9.4% ionized. Using this information, calculate Ka for HA.
slavikrds [6]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Let's solve ~

Initial concentration of weak acid HA = 0.200 M

and dissociation constant ({ \alpha}) is :

\qquad \sf  \dashrightarrow \:  \alpha =  \frac{dissociation \:  \: percentage}{100}

\qquad \sf  \dashrightarrow \:  \alpha =  \frac{9.4}{100}  = 0.094

Now, at initial stage :

  • \textsf{ Conc of HA = 0.200 M}

  • \textsf{Conc of H+ = 0 M}

  • \textsf{Conc of A - = 0 M}

At equilibrium :

  • \textsf{Conc of HA = 0.200 - 0.094(0.200) = 0.200(1 - 0.094) = 0.200(0.906) = 0.1812 M}

  • \textsf{Conc of H+ = 0.094(0.200)  = 0.0188 M}

  • \textsf{Conc of A - = 0.094(0.200)  = 0.0188 M}

Now, we know :

\qquad \sf  \dashrightarrow \: { K_a = \dfrac{[H+] [A-]}{[HA]}}

( big brackets represents concentration )

\qquad \sf  \dashrightarrow \: { K_a = \dfrac{0.0188×0.0188}{0.1812}}

\qquad \sf  \dashrightarrow \: { K_a = \dfrac{0.00035344}{0.1812}}

\qquad \sf  \dashrightarrow \: { K_a \approx 0.00195 }

\qquad \sf  \dashrightarrow \:  {K_a \approx 1.9 × {10}^{-3} }

7 0
2 years ago
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