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True [87]
3 years ago
5

You are writing a report on whether global warming really exists. Which of the following resources would be the most scientifica

lly reliable? (2 points) an article containing documented experiments and data a newspaper article quoting leaders of several countries a website with articles posted by people around the world an article containing a poll taken of students at leading universities
Chemistry
1 answer:
Ierofanga [76]3 years ago
5 0

For this question, I would definitely say an article containing documented experiments and data because if you are talking about something to do with science, you should use science to prove your point. People usually believe articles done by someone who did the experiment themselves rather than quoting other people from their findings.

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If 45.0 mL of ethanol (density =0.789g/mol) initially at 6.0°C mix with 45.0 mL of water (density =1.0 g/mol) initially at 28.0°
Likurg_2 [28]

The final temperature of the mixture : 21.1° C  

<h3>Further explanation  </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released  

Q in(gained) = Q out(lost)  

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Q ethanol=Q water

mass ethanol=

\tt mass=\rho\times V\\\\mass=0.789\times 45=35.505~g

mass water =

\tt mass=1~g/ml\times 45~ml=45~g

then the heat transfer :

\tt 35.505\times 2.42~J/g^oC\times (t-6)=45\times 4.18~J/g^oC\times (28-t)\\\\85.922t-515.533=5266.8-188.1t\\\\274.022t=5782.33\rightarrow t=21.1^oC

5 0
2 years ago
How much energy does an X-ray with an 8 nm (8 x 10-9 m) wavelength have?
Vlada [557]

Answer: 2.48×10^-17 J

Explanation:

Given the following :

Wavelength = 8nm (8 x 10^-9 m)

Energy(e) of X-ray =?

Energy=[speed of light(c) × planck's constant (h)] ÷ wavelength

Speed of light = 3×10^8m/s

Planck's constant = 6.626×10^-34 Js

Wavelength = 8 x 10^-9 m

Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9

Energy = [19.878×10^(8-34)] / 8 x 10^-9

Energy = 2.48475 × 10^(-26+9)

Energy = 2.48×10^-17 J

8 0
2 years ago
Which property determine am atoms ability to attract electrons shared in a chemical bond?
brilliants [131]
Hello there!

Electronegativity is what determine's an atoms ability to attract electrons shared in a chemical bond.Ionization, atomic radius, and also <span> ionic radius both would not determine this as they wouldn't have any similar bond that would attract.
</span><span>
Your correct answer would be (option c)

</span><span>A. ionization 

B. atomic radius

C. electronegativity

D. ionic radius

I hope this helps you!</span>
8 0
3 years ago
Two moles of magnesium and five moles of oxygen are placed in a vessel. When magnesium is ignited ca two moles of magnesium and
Greeley [361]
Limiting reactant in this experiment would be Magnesium since it will run out first
4 0
3 years ago
. a large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is sub
elena-s [515]

The large piece of jewelry  that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml

To solve this problem the formulas and the procedures that we have to use  are:

  • v = v(f)-v(i)
  • d = m/v

Where:

  • d= density
  • m= mass
  • v= volume
  • v(f) = final volume
  • v(i) = initial volume

Information about the problem:

  • m = 132.6 g
  • v(i) = 48.6 ml
  • v(f) = 61.2 ml
  • v = ?
  • d =?

Applying the volume formula we get:

v = v(f)-v(i)

v = 61.2 ml - 48.6 ml

v = 12.6 ml

Applying the density  formula we get:

d = m/v

d = 132.6 g/12.6 ml

d = 10.523 g/ml

<h3>What is density?</h3>

It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.

Learn more about density in: brainly.com/question/1354972

#SPJ4

3 0
1 year ago
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