I need you help in this homework

Carbon dioxide and water vapor are variable gases because

ehidna [41]

Answer:

Their amount is not consistent at all parts of the atmosphere

Explanation:

Carbon dioxide and water vapor exists as variable gases in the atmosphere because their amount varies from one part to the other.

The abundance of variable gases is influenced strongly by activities from the surface.
This variability occurs a result of the contribution from from life activities on the surface.
Also, due to properties of the gases, they are not evenly distributed.
At certain altitudes, the amount of water vapor can be low.

7 0

3 years ago

The nucleus of a fluorine atom has a charge of? A. +19 B. 0 C. +9 D. +1

posledela

Answer:

A fluorine atom has nine protons and nine electrons, so it is electrically neutral. If a fluorine atom gains an electron, it becomes a fluoride ion with an electric charge of -1.

7 0

3 years ago

Consider the balanced chemical reaction below. When the reaction was carried out, the calculated theoretical yield for carbon di

jeka94

Answer:

Percent Yield = 94.237%

Explanation:

CO = Carbon Dioxide = Molar Mass 28g/mol

C = Carbon = 12g/mol

O = Oxygen = 16g/mol

Theoretical yield = 93.7 grams

Actual yield = 88.3 grams

Percent yield = (actual yield /theoretical yield )x100

Percent Yield = (88.3/93.7)x100

Percent Yield = 94.237%

8 0

2 years ago

Read 2 more answers

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

1.

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

$pOH=-\log[OH^-]$

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$ :

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

3 0

3 years ago

If 185 mg of acetaminophen were obtained from a tablet containing 350 mg of acetamino- phen, what would be the weight percentage

Alex73 [517]

Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.

Amount of acetaminophen initially taken = 350 mg

Amount of acetaminophen obtained after recovery =185 mg

$Weight percentage recovery =\frac{mass recovered}{mass originally taken}*100$

= $\frac{185 mg}{350 mg}*100$

= 52.9%

8 0

3 years ago