Biochemistry which studies the chemical composition, structure, and the reaction of compounds found in non-living organisms.

List in ascending order the different layers of atmosphere, their heights, and main their features. Help please!!! :•)

Anika [276]

Answer:

Troposphere

Stratosphere

Mesosphere

Thermosphere

Exosphere

Explanation:

These are all listed in ascending order, hope this helps!

6 0

3 years ago

Is a forward or reverse reaction favored?

Alex Ar [27]

Answer:

(a) $CH_3COOH \rightleftharpoons CH_3COO^-+ H^+$ : reverse reaction is favored.

(b) $AgCl \rightleftharpoons Ag^+ + Cl^-$ : reverse reaction is favored.

(c) $Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-$ : reverse reaction is favored.

(d) $A+B\rightleftharpoons C$ : forward reaction is favored.

Explanation:

Hello,

(a) $CH_3COOH \rightleftharpoons CH_3COO^-+ H^+$ :

In this case, since the Ka is lower than 1, we infer the reverse reaction is favored since the reactant (acetic acid) will tend to have a higher concentration.

(b) $AgCl \rightleftharpoons Ag^+ + Cl^-$ :

In this case, since the Ka is lower than 1, we infer the reverse reaction is favored since the reactant (silver chloride) will tend to have a higher concentration.

(c) $Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-$

In this case, since the Ka is lower than 1, we infer the reverse reaction is favored since the reactant (aluminium hydroxide) will tend to have a higher concentration.

(d) $A+B\rightleftharpoons C$

In this case, since the Ka is greater than 1, we infer the forward reaction is favored since the product (C) will tend to have a higher concentration.

Regards.

6 0

3 years ago

Identify ways to reduce the risk of liquid bumping while heating.

postnew [5]

The most common way of preventing bumping is by adding one or two boiling chips to the reaction vessel. However, these alone may not prevent bumping and for this reason it is advisable to boil liquids in a boiling tube, a boiling flask, or an Erlenmeyer flask.

7 0

2 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago

Someone please help me:

Stels [109]

Answer: The number to the left of AC should be 6.

Explanation: The balanced chemical reaction is one in which the number of atoms of each element on the reactant side must be equal to the number of atoms on product side.

The given equation $A_2B + DC_3 \rightarrow AC + D_2B_3$ is unbalanced as the atoms on the reactant side are not same as number of atoms on product side. This equation is called as skeletal equation.

The balanced chemical equation is :

$3A_2B + 2DC_3 \rightarrow 6AC + D_2B_3$

Thus the number to the left of AC is 6.

6 0

2 years ago

Read 2 more answers