Chemistry, 03.12.2019 05:31 sativataurus

Which two processes are responsible for keeping the percentage of atmospheric oxygen at relatively constant levels?

Answers: 1

Show answers

Answers

I believe the two processes which the question is talking about is photosynthesis and cellular respiration. These processes are responsible for keeping an organism alive and living. Photosynthesis makes so the plant can live off of sunlight, water, and oxygen. Cellular respiration make so that nutrient and oxygen transport can occur.

Answer from: micahsocool23

Photosynthesis and respiration are responsible for it

Answer from: Quest

when the volume of gas is changed from 717.73 ml to 852 ml the temperature will change from 315 c to 452 c

calculation

this is calculated using the charles law formula

that is, v1/t1 = v2/t2

v1=?

t1 = 315 c into kelvin = 315 +273 = 588 k

v2 = 852 ml

t2 = 425c + 273 =698 k

make v1 the subject of the formula ( by multiplying both side by t1)

v1=( v2 xt1) / t2

v1= (852 ml x 588 k)/ 698 k= 717.73 ml

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Another question on Chemistry

Chemistry, 21.06.2019 22:30

In order to calculate the amount of heat transferred you must know the __ and specific heat of the material, as well as the change in temperature. a. volume b. density c. mass d. enthalpy

Answers: 1

Answer

Chemistry, 21.06.2019 22:30

Check the correct box to describe the periodic trends in electronegativity. electronegativity across a period: decreases. increases. electronegativity down a group: decreases. increases.

Answers: 2

Answer

Chemistry, 22.06.2019 05:30

Describe the interaction that occurs between two objects with the same electrical charge.

Answers: 1

Answer

Chemistry, 22.06.2019 11:20

Sodium nitrite (nano2) reacted with 2−iodooctane to give a mixture of two constitutionally isomeric compounds of molecular formula c8h17no2 in a combined yield of 88%. draw reasonable structures for these two isomers. click the "draw structure" button to launch the drawing utility. place the two compounds in the appropriate boxes below.

Answers: 1

Answer

You know the right answer?

Which two processes are responsible for keeping the percentage of atmospheric oxygen at relatively c...

Questions