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xz_007 [3.2K]
3 years ago
13

Scientists who are investigating what has happened over the years to the people who were exposed to radiation during the 1986 Ch

ernobyl nuclear disaster are conducting what type of study
Chemistry
1 answer:
aleksandrvk [35]3 years ago
8 0

Answer:

Observational study

Explanation:

An investigation of what happened over the years to the people who were exposed to radiation during the 1986 Chernobyl nuclear disaster would be termed <u>an observational study</u> because it will require observing subjects and finding out if there is a kind of correlating factor in the subjects that could indicate the level of exposure to the nuclear radiation.

There are basically two types of research studies:

1. Observational studies

2. Experimental studies

Observational studies involve just the observation of subjects and trying to find out some correlation factors. This kind of study could be prospective when it deals with what could happen in the future based on certain factors or retrospective when it considers what has already happened. An observational study can be cross-sectional, case-only, case-control, or cohort study.

Experimental studies are also referred to as randomized or controlled trial studies. Here, subjects are grouped and each group receives a different treatment in order to isolate the effect of a particular factor or variable relating to the treatment.

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A student conducting the iodine clock experiment accidentally makes an s2o32- stock solution that is too concentrated. how will
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Ede4ka [16]

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I and II      I think

Explanation:

4 0
2 years ago
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Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50
devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C_{V} You can work out C_{V}

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC_{V} ∆T

Polyatomic gas: C_{V} = 3R

∆U= nC_{V} ∆T

∆U= 28g x C_{V} x (350K - 58.3K)

∆U = 28C_{V} x 291.7

∆U = 10967.6 x C_{V}

5 0
3 years ago
A rectangular block has the following dimensions: 3.21 dm, 5.83 cm, and 1.84 in. The block has a mass of 1.94 kg. What is the de
spin [16.1K]

The density of the rectangular block in g/mL is 7.0.

<u>Given the following data:</u>

  • Mass of block = 22.8 gra1.94 kg
  • Length of block = 3.21 cm
  • Width of block = 5.83 cm
  • Height of block = 1.84 in.

To find the density of the block in g/mL:

First of all, we would determine the volume of the rectangular block by using the following formula:

Volume = length × width × height

<u>Conversion:</u>

1 in = 2.54 cm​

5.83 in = X cm

Cross-multiplying, we have:

X = 2.54(5.83)\\\\X = 14.81 \; cm

Volume = 3.21 × 5.83 × 14.81

Volume = 277.16 cubic centimeters.

<u>Note</u>: Milliliter (mL) is the same as cubic centimeters.

1000 grams = 1 kg

Y grams = 1.94 kg

Cross-multiplying, we have:

Y = 1940 grams

Now, we can find the density:

Density = \frac{Mass}{Volume}\\\\Density = \frac{1940}{277.16}

<em>Density </em><em>= 7</em><em>.0 g/mL</em>

Therefore, the density of the rectangular block in g/mL is 7.0.

Read more: brainly.com/question/18320053

4 0
2 years ago
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