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Rom4ik [11]
3 years ago
15

Please help (: What are the sources and stages of a river ?

Chemistry
2 answers:
krek1111 [17]3 years ago
7 0

Water, ALSO if you cant figure out the answer looking up on g00gl3e

defon3 years ago
5 0

3 Stages of a River

YOUTHFUL STAGE (UPPER COURSE) – V- Shaped Valley > Erosion. ...

MATURE STAGE (MIDDLE COURSE) – Meanders > Erosion and Deposition.

OLD AGE STAGE (LOWER COURSE) – Floodplains > Deposition. ...

Advantages. Scenic Attraction. ...

Dangers. Flooding – Damage to property, land, animals and homes. ...

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Write down five things that you personally could do in your day-to-day life to help with the problem of climate change. Write do
beks73 [17]
Before that, we need to know what causes climate change. One of the main reasons are the carbon dioxide released into the atmosphere, it traps heat but allow more heat to enter the earth, therefore climate increases over the years.

1. Set air conditioner to a higher temperature (eg. 25°C) or turn on fans instead.
Operating AC can use a lot of energy, and these energy is produced by burning if fossil fuels which releases carbon dioxide.

2. Take public transports instead of cars. The more private cars there are, the more pollutants it release, these pollutants includes carbon dioxide.

3. Turn off all the lights / electrical appliances when we leave a place. Again, electrical appliances require energy which is produced by burning of fossil fuels that gives out carbon dioxide.

4. Recycle and reuse paper that is only used on one side. Papers are extracted from trees and one main function of trees is to absorb carbon dioxide from the air. If many trees are cut, the amount of carbon dioxide in air would increase.

5. Do the laundry only if you have clothes full for the machine. If the washing machine is not full and you're starting it up, it really uses a lot of energy and water, which all again it triggers the release of carbon dioxide.
4 0
4 years ago
out of 2.00 50 mL of water is added to 30 mL of a 0.50 M salt solution. What is the new concentration? (To write your answer usi
vodka [1.7K]

Answer:

The concentration of the new solution will be 0.31 M

Explanation:

The number of moles per liter of the solution is 0.50 mol. Then, in 50 ml there will be:

50 ml · (0.50 mol / 1000 ml) = 0.025 mol.

If we add 30 ml and assuming that the solution is an ideal solution, the final volume will be 80 ml.

Then, 0.025 mol will be present in 80 ml solution. In 1 l there will be:

1000 ml · (0.025 mol / 80 ml) = 0.31 mol.

The concentration of the new solution will be 0.31 M.

3 0
3 years ago
quizlet a basement is successfully sealed so that more radon may not enter the space, but 5.7×107 radon atoms are already trappe
12345 [234]

After 25 days, it remains radon 5.9x10^5 atoms.

Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.

N(Ra) = 5.7×10^7; initial number of radon atoms

t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days

n = 25 days / 3.8 days

n = 6.58; number of half-lifes of radon

N1(Ra) = N(Ra) x (1/2)^n

N1(Ra) = 5.7×10^7 x (1/2)^6.58

N1(Ra) = 5.9x10^5; number of radon atoms after 25 days

The half-life is independent of initial concentration (size of the sample).

More about half-life: brainly.com/question/1160651

#SPJ4

6 0
2 years ago
Correct lewis dot structure for chlorine (Cl)?
shepuryov [24]
Answer with pictures

5 0
3 years ago
An isolated atom of a certain element emits light of wavelength 655 nm when the atom falls from its fifth excited state into its
dangina [55]

Answer:

\frac{1}{\lambda_{6-5}} \approx 752nm

Explanation:

From the question we are told that

Light wavelength  \lambda_l=655nm

Light wavelength atom fall x_L=5th-2nd

Photon wavelength  \lambda_p=435nm

Photon wavelength atom fall x_P=^th-2nd

Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by

\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )

Therefore for initial drop of 5th to 2nd

\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )

Therefore for initial drop of 6th to 2nd

\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )

Generally we subtract (5th to 2nd) from (6th to 2nd)

\frac{1}{\lambda_{5-2}}-   \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )

\frac{1}{\lambda_{5-2}}-   \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )

\frac{1}{\lambda_{5-2}}-   \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}

\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}

\frac{1}{\lambda_{5-6}}=1.33*10^{-3}

Therefore for 6th to 5th stage is mathematically given by

\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}

\frac{1}{\lambda_{6-5}}=751.879nm

\frac{1}{\lambda_{6-5}} \approx 752nm

6 0
3 years ago
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