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Alisiya [41]
3 years ago
13

5. Why doesn't the air at a frontal boundary mix?​

Chemistry
1 answer:
Vsevolod [243]3 years ago
4 0

Answer:

At a front, the two air masses have different densities, based on temperature, and do not easily mix. One air mass is lifted above the other, creating a low pressure zone.

Explanation:

Hope this helps!

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In the following compound (HCl) How many electrons are gained and lost by each atom?
Verizon [17]

Explanation:

In HCL, one positive atom is given to chlorine from hydrogen so that it can complete it's octate. chlorine take one electron from hydrogen.

In NaCl, Sodium takes one electron from chlorine to complete its orbit with eight electrons. Chlorine gives one electron to sodium.

8 0
3 years ago
Please help me with this three questions
zheka24 [161]

Answer:

1. A. Thyroid Gland Function

B. detecting Blood Clots

C. Treating Cancer

2. An example would be how many are used as tracers for diagnostic purposes

3. Beneficial because it gives us ways to eleminate pests, by the sterile insect technique

Number three is kinda iffy a bit difficult to explain it may be explained better in your book.

7 0
3 years ago
The vapor pressure of benzene is 100.0 mmhg at 26.1°c. calculate the vapor pressure of a solution containing 28.2 g of camphor (
FrozenT [24]
<span>Answer: Moles C10H16O = 24.6 g/ 153.23 g/mol = 0.161 Moles benzene = 98.5 g / 78.1121 g/mol = 1.26 Mole fraction benzene = 1.26 / 1.26 + 0.161 = 0.887 vapor pressure = 100 x 0.887 = 88.7 mm Hg</span>
5 0
3 years ago
Base your answers to questions 74 through 76 on the data table below and on your knowledge of Earth
Serjik [45]

Answer:

See explanation

Explanation:

Now we have, the graph attached.the stable disintegration product of C-14 is N-14.

Then;

Since the mass of C-14 originally present is 64g, at a time t= 17100 years, we will have;

N/No = (1/2)^t/t1/2

N = mass of C-14  at time t

No= mass C-14 originally present

t = time taken for  N amount of C-14 to remain

No = mass of C-14 originally present

t1/2 = half life of C-14

N/64 = (1/2)^17,100/5730

N/64 = (1/2)^3

N/64 = 1/8

8N = 64

N = 8 g

Download xlsx
6 0
3 years ago
Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read
lapo4ka [179]

Answer:

1) 18.91 mL

2) 2.35 %

Explanation:

1. The volume of base that was required is equal to the <em>difference between the reading at the end of the titration and the reading at the beginning</em>:

  • V = 20.95 mL - 2.04 mL = 18.91 mL

2. The mass percent can be written as:

  • Mass Percent = Mass solute / Total mass * 100%

For this problem:

  • Mass solute = 2.28 g
  • Total mass = 96.92 g

We<u> input the data and calculate the mass percent</u>:

  • % mass = 2.28/96.92 * 100% = 2.35 %
7 0
3 years ago
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