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dedylja [7]
3 years ago
11

Matter that emits no light at any wavelength is called

Physics
2 answers:
Tpy6a [65]3 years ago
4 0

The answer is D.dark matter

ruslelena [56]3 years ago
3 0
<span>A negative space is the space around and between the real object. It is the space that forms around the subject after it has altered to give an artistic feel. A dark energy is hypothesized to infuse all of space accelerating the expansion of the universe. It is also an unknown form of energy. A black hole exhibits a strong gravitational pull that no astronomical object can escape from it. A dark matter is a black space in the universe that constitutes most of the unknown. They so not emit light at any wavelength. The answer is letter D.</span>
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ad-work [718]

Answer:the answer is true

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Hope this helped

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Which physical activity would a global positioning system resource help you with?
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The physical activity that a global positioning system resource can help me with is : ( A )  Hiking

<h3>Function of Global positioning system ( GPS ) </h3>

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Learn more about GPS : brainly.com/question/9795929

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4 0
2 years ago
If Star A and Star B have the same absolute magnitude, but Star A is brighter, what does that tell us?
Hatshy [7]
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4 0
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Read 2 more answers
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
3 years ago
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