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Scrat [10]
3 years ago
5

Describe how greater distance affects the appearance of an object’s size and brightness

Chemistry
1 answer:
jeka943 years ago
4 0

Answer:

Distance also affects the apparent brightness of an object. A distant, but very energetic star can appear dimmer to us than a lower-energy, but closer one. ... Each of their effective temperatures results in a different brightness for the object. The quasar, however, is very far away, and so appears dim.

Explanation:

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One of the steps in processing some metals is smelting. what does smelting do for a metal?
vichka [17]
Hey there,
Smelting's main use is to produce a base metal from its ore

Hope this helps :))

~Top
8 0
3 years ago
Plutonium-240 decays according to the function \[Q(t)=Q _{o}e ^{-kt}\] where Q represents the quantity remaining after t years a
marishachu [46]
To answer the question above, substitute the given values to the given equation,
                                   Q(t) = Q x e^-kt
  
                          12 grams = (36 grams) x e^(-0.00011)(t)

Solving for t gives t = 9,987.38 years or approximately equal to 9,990 years. Thus, the answer is letter C.












4 0
3 years ago
Read 2 more answers
Rocks are classified based on
Viefleur [7K]
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7 0
3 years ago
I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti
Tanya [424]
<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
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[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
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pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
4 0
3 years ago
What volume is needed to store 0.80 moles of helium gas at 204.6 kpa and 300 k?
shepuryov [24]
Answer is: 9,7 L is needed to store helium gas.
n(He) = 0,80 mol.
p(He) = 204,6 kPa.
T = 300 K.
R = 8,314 J/K·mol; universal gas constant.
Use ideal law eqaution: p·V = n·R·T.
V = n·R·T / p.
V(He) = 0,80 mol · 8,314 J/K·mol · 300 K ÷ 204,6 kPa.
V(He) = 9,75 L.


3 0
3 years ago
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