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nikdorinn [45]
3 years ago
14

The average person in the United States is exposed to the following amount of radiations annually. Rank the following source of

radiation in the increasing order of their ability to cause harm to living tissue.
Weapons-test fallout 1 millirem
Cosmic radiation 26 millirems
Air (radon-222) 0.198 rems
Diagnostic X rays 0.040 rems
Television tubes 11 millirems
Nuclear medicine 0.015 rems
Ground 33 millirems
Chemistry
2 answers:
Papessa [141]3 years ago
8 0
<span>#1 is air radon, #2 is x-ray, #3 is ground, #4 is cosmic radiation, #5 is TV tube, #6 is weapons test fallout . That's all I got hope I helped!</span>
Nookie1986 [14]3 years ago
6 0

Answer:

The increasing order of the ability to cause harm to living tissue:

Weapons  < Television tubes < Nuclear medicine < Cosmic radiation <Ground < Diagnostic X rays < Air (radon-222)

Explanation:

Here, the ability of the radiation to harm living tissue is measured in terms of millirem

Unit conversion:

1\ millirem = \frac{1}{1000}\  rem = 0.001\ rem\\\\or,\ 1\ rem = 1000\ millirem

The given radiation doses can be converted into units of millirems for comparison:

Weapons test fallout = 1 millirem

Cosmic radiation = 26 millirems

Air (radon-222) =  0.198 rems  = 198 millirems

Diagnostic X rays =  0.040 rems  = 40 millirems

Television tubes = 11 millirems

Nuclear medicine =  0.015 rems  = 15 millirem

Ground = 33 millirems

Since a greater dosage of radiation would lead to greater risk, the increasing order of their ability to cause harm to living tissue would be:

Weapons  < Television tubes < Nuclear medicine < Cosmic radiation <Ground < Diagnostic X rays < Air (radon-222)

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A concentrated salt solution has a mass of 5.222 g for a 5.000 mL sample.
ad-work [718]

The specific gravity or relative density of a substance is the ratio of its density to the density of a reference material. The relative density of the concentrated salt solution is 1.044.

Mathematically;

Density of the concentrated salt = mass of salt/volume of salt = 5.222 g/5.000 mL = 1.044 g/mL

In the case of specific gravity, the reference material is always water and water has a density of 1 g/mL.

Hence, specific gravity of the concentrated salt solution =

Density of concentrated salt solution/density of equal volume of water

= 1.044 g/mL/1 g/mL

= 1.044

Note that specific gravity is dimensionless.

Learn more: brainly.com/question/9638888

6 0
3 years ago
1) What mass of Na2CO3 is required to make 50cc of its seminormal solution?
love history [14]

Answer:

m=1.325gNa_2CO_3

Explanation:

Hello,

In this case, by considering the given seminormal solution, we infer it is a 0.5-N solution which means that we can obtain the equivalent grams as shown below for the 55 cc (0.055 L) volume:

eq-g=0.5eq-g/L*0.050L=0.025eq-g

Next, since sodium carbonate has two sodium ions with a +1 oxidation state each, we can obtain the moles:

mol=0.025eq-gNa_2CO_3*\frac{1molNa_2CO_3}{2eq-gNa_2CO_3}\\ \\mol=0.0125molNa_2CO_3

Finally, the mass is computed by using its molar mass (106 g/mol)

m=0.0125molNa_2CO_3*\frac{106gNa_2CO_3}{1molNa_2CO_3} \\\\m=1.325gNa_2CO_3

Regards.

7 0
3 years ago
He substance, cocl2, is useful as a humidity indicator because it changes from pale blue to pink as it gains water from moist ai
Anettt [7]

Answer:

Cobalt chloride

Explanation:

It is a molecule formed of one Cobalt atom, and two Chlorine atoms.

5 0
3 years ago
The molar mass of I2 is 253.80 g/mol, and the molar mass of NI3 is 394.71 g/mol. How many moles of I2 will form 3.58 g of NI3?
Nat2105 [25]
To answer the problem above first we need to find the difference of molar mass of NI3 from I2, 394.71 g/mol - 253.80 g/mol = 140.91 g/mol. Knowing the molar mass of the difference of NI3 from I2, in equation mass (g) / moles (mol) = molar mass, then we substitute. 3.58g / moles = 140.91 g/mol.
moles = 3.58 / 140.91 = 0.025 moles.
6 0
4 years ago
2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume
Cerrena [4.2K]
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
5 0
3 years ago
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