CHEMISTRY COLLEGE

Choose all the answers that apply. Which of the following is true about the sun's radiation? A) It is long-wave.
B) It is short-wave.
C) 40 percent is reflected by the atmosphere.
D) 60 percent reaches Earth's surface.
E) None of the rays are harmful.

Answers

Answer 1
Answer: B. It is shortwave.....
Answer 2
Answer:

Answer:

Your answers are B,C, and D

Explanation:

Because a and e are wrong it is these three.

Edit pls mark brainliest

Related Questions

Write Lewis structural formulas for the following molecules: NH3 SO2 CH3OH HNO2 N2 CH2O
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You need to prepare an acetate buffer of pH 5.47 from a 0.809 M acetic acid solution and a 2.20 M KOH solution. If you have 680 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.47 ? The pa of acetic acid is 4.76. Be sure to use appropriate significant figures.
22. What is the molar mass of oxygen (O2)?A- 15.9994 g/molB- 1.204 x 1024 g/molC- 6.02 x 1023 g/molD- 31.9988 g/mol
A water wave has a frequency of 10 Hz and a wavelength of 150 m. What is the wave speed? *Equation to use: wave speed (m/s) = frequency (Hz) x wavelength (m)*

The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the configurations of the oxirane products, A and B are Product A: _______ Product B: _______ Will either of these two oxirane products rotate the plane of polarization of plane polarized light

Answers

Answer:

Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane.

Explanation:

A double bond is converted to an oxirane through oxidation by peracids e.g. mCPBA (meta-chloroperoxybenzoic acid).

Epoxidation can occur at both face of double bond result in formation of two stereoisomers.

Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane

Both A and B contain plane of symmetry. Hence, both the products are achiral. So, they do not rotate the plane of polarization of plane polarized light.

You just measured a metal cylinder and obtained the following information: mass - 3.543 g diameter -0.53 cm height = 4.40 cm. 265 • Determine the volume (V). (V=nrºh, where r = radius, h = height, T - 3.14) V= (3.14 36205² 4.402.17cm? • Determine density of the cylinder D. 3. SMS-365923 197

Answers

The density of the cylinder would be 3.652 gram/ cm³

.

What is density?

It can be defined as the mass of any object or body per unit volume of the particular object or body. Generally, it is expressed as in gram per cm³ or kilogram per meter³.

As given in the problem, you just measured a metal cylinder and obtained the following information: mass - 3.543 grams, diameter 0.53 cm, height = 4.40 cm , and we have to calculate the density of the cylinder,

mass of the cylinder = 3.543 grams

the volume of the cylinder = πr²h

                                            = 3.14 ×.265²×4.4

                                            =0.97 cm³

By using the above formula for density

ρ = mass of the cylinder/volume of the cylinder

  = 3.543 grams/0.97 cm³

  =3.652 grams/ cm³

Thus,the density of the cylinder would be 3.652 grams/ cm³.

To learn more about density from here, refer to the link;

brainly.com/question/15164682

#SPJ2

Answer:

V cylinder = \pi r^2 h = \pi(0.53 cm/2)^24.4cm= 0.97 cm^3

note that r=diameter/2

density = m/V= 3.543/0.97= 3.65 g/cm^3

100.0 g of liquid copper (molar mass 63.546 g/mol; melting point 1358 K; density 8.02 g/mL) is placed in a rigid container of volume 10.0 L at temperature 1508 K. The container is placed in an evacuated chamber and a small hole of area 3.23 mm2 is made in the upper container wall. After 2.00 hours, the mass of copper in the container has decreased by 1 0.0168 g. Assuming the mass loss is due to effusion, calculate the vapor pressure of liquid copper at 1508 K. Hint: because the liquid constantly evaporates, the pressure inside the container is constant

Answers

Answer:

8.912x10^-18

Explanation:

-dn/dt = pANa/2piMRT

100 g = initial copper

Number of moles = 100/63.546

= 1.5736

Mass of copper left = 100-10.0168

= 89.9832

Moles = 89.9832/63.546

= 1.4160

dn = 1.4160-1.5736

= -0.1576

dt = 2 hrs

A = 3.23mm² = 3.23x10^-6

M = 63.546

T = 0.0821

T = 1508k

Na = 6.023x10²³

When we insert all these into the formula above

We get

P = 8.912x10^-18atm

For each of the following redox reactions in the, identify the oxidizing agent and the reducing agent.2Na(aq)+2H2O(l)→2NaOH(aq)+H2(g)
C(s)+O2(g)→CO2(g)
2MnO−4(aq)+5SO2(g)+2H2O(l)→2Mn2+(aq)+5SO2−4(aq)+4H+(aq)

Answers

anywhere between 5sogt2 and 776sogt2

How much energy is required to heat the solid iron from 1535 ∘C to 2000 ∘C?

Answers

"206.46 J/g" energy required to heat the solid iron.

Energy due to change in temperature:

The movements or the activity of microscopic particles known as atoms, molecules, or ions in solids, solvents, as well as gases.

Thermal energy may go from one item towards the another one.

Given:

Temperature, T_1 = 1535°C

                       T_2 = 2000°C

Change, \Delta T = T_2-T_1

We know the relation,

→ Q = mC_p \Delta T

By putting the values,

      = 0.444(2000-1535)

      = 206.46 J/g

Thus the answer above is right.

Find out more information about energy here:

brainly.com/question/2033877

Answer:

Q=206.46(J)/(g)

Explanation:

Hello,

In this case, since the energy due to the change of temperature is computed via:

Q=mCp\Delta T

Thus, since the specific heat of iron is 0.444 J/(g°C), the heat per unit of mass turns out:

Q=0.444(J)/(g\°C)(2000-1535)\°C\n \nQ=206.46(J)/(g)

Best regards.

What action is NOT necessary before lighting a Bunsen burner?A)Make sure there are no holes in the hose connected to the burner.
B)Leave the gas stop open for a few minutes before lighting to clear the line.
C)Clear the burner area of flammable materials such as solvents and papers.
D)Check that the gas regulator and air inlet on the burner can be adjusted.

Answers

The Bunsen burner is a heating apparatus used to heat samples in the laboratory. The correct answer is option B: )Leave the gas stop open for a few minutes before lighting to clear the line.

A Bunsen burner is commonly used as a heating apparatus. It contains a gas that can be controlled using a gas regulator.

It is necessary to do the following before lighting the Bunsen burner;

  • Make sure there are no holes in the hose connected to the burner.
  • Clear the burner area of flammable materials such as solvents and papers.
  • Check that the gas regulator and air inlet on the burner can be adjusted.

So, it is not necessary to Leave the gas stop open for a few minutes before lighting to clear the line before lighting a Bunsen burner.

Learn more: brainly.com/question/743920

Answer:

the answer is opetuon B
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