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vladimir1956 [14]
3 years ago
13

What is being done to reduce the levels of acid rain?

Chemistry
1 answer:
ludmilkaskok [199]3 years ago
8 0
<span>EPA’s Acid Rain Program</span>
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Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

Therefore,  \lambda = 1 / 6580.62 = 0.000152 m

(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

#SPJ4

8 0
1 year ago
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP
Dvinal [7]
Vol.250 before its to much pressure

7 0
3 years ago
Read 2 more answers
Sperm can be carried by _____ grains. ANSWERSSSSSSSSSSSSSSSSS O spore O stamen O pollen
7nadin3 [17]

What is pollen in the reproduction cycle of flowering plants?

A pollen grain is a microspore containing the male gametophyte, usually reduced to two undivided cells, each with one haploid (n) nucleus. These cells are surrounded by a very resistant wall, the exine, which generally has apertures, zones with less resistance which will allow the germination of the pollen tube.

Explanation of the reproduction cycle (cf diagram above)

A given species produces flowers bearing stamens. According to species, these flowers can be unisexual (stamens only) or bisexual (stamens and carpels).

The stamen anthers include 4 pollen sacs containing sporogenous cells (diploid=2n). These sporogenous cells undergo meiosis, each producing 4 microspores (haploid=n). Two nuclei are then formed by mitosis : the vegetative nucleus and the generative nucleus. The latter divides to form 2 sperms. Simultaneously the wall of the microspores becomes thicker and takes the characteristic shape of the species : it is a pollen grain (see: What are the morphological characteristics of pollen and spore grains?). In the majority of species, the 4 grains (resulting from the 4 microspores) split up into single grains; in some cases, they remain together (tetrad = group of 4 grains). When mature, pollen grains are released by the opening of the anthers.

A pollen grain is aimed at reaching another flower of the same species, bearing carpels. The ovaries contain ovules, in which meiosis occurs, then mitoses. It results in an embryo sac with 8 nuclei, among which an egg cell and 2 central cells. When a pollen grain arrives on another flower (see : How are the spores and pollen grains transported?), it is received by the stigmas.

The pollen grain germinates through an opening of the wall: the vegetative nucleus develops into a pollen tube which is guided by the style to the ovary, then enters the micropyle of an ovule. The pollen tube releases 2 sperm nuclei into the ovule: one of the sperms fuses with the egg cell into a zygote (2n), while the other sperm fuses with central nuclei and gives rise to albumen (= food source). There are generally several ovules in an ovary : each one can be fertilized by a distinct pollen grain.

Each fertilized ovule and its albumen form a seed that will develop into a new individual of this species. hope it works

4 0
3 years ago
Read 2 more answers
You have added IgG-sensitized red cells to the negative indirect antiglobulin test result of the antibody screen procedure. You
Ivanshal [37]

Answer:

Positive

Explanation:

The antibody screen interpretation is positive. That's because of clump formation when antibody reacts with the antigen and agglutination occurs.

It means that the antibodies were present in the sample that react with the antigen present in the test tube.

5 0
3 years ago
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0
3 years ago
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