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Xelga [282]
3 years ago
5

If we know the specific heat of a material, can we determine how much heat is released under a given set of circumstances?

Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
6 0

Answer:

Yes

Explanation:

literally have no idea i just got it right on ck-12 lol

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15. If the moon is 384,400km away. Does it take more than a second for light to travel from the Earth to the moon?​
marishachu [46]

Answer:

Yes

Explanation:

The speed of light is about 300 000 km per second, so light takes about 1.28 seconds to travel from the Moon to the Earth.

3 0
1 year ago
Which option describes energy being released as heat?
Mnenie [13.5K]

Answer:

I’m pretty sure it’s Lions sleeping after a big meal

Explanation:

8 0
4 years ago
Read 2 more answers
If 1.00 g of a hydrocarbon is combusted and found to produce 3.14 g of co2, what is the empirical formula of the hydrocarbon?
photoshop1234 [79]
The combustion reaction is as expressed,

                CxHy + O2 --> CO2 + H2O

The mass fraction of carbon in CO2 is 3/11. Hence,
       mass of C in CO2 = (3.14 g)(3/11) = 0.86 g C.

Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g. 

     moles of C = 0.86 g C / 12 g = 0.0713
     moles of H = 0.14 g H / 1 g  = 0.14

The empirical formula for the hydrocarbon is therefore, CH₂.
7 0
3 years ago
Question 14 of 17
artcher [175]

Since Lutetium-177 is a beta and gamma emitter, the daughter nuclide produced from the decay of this radioisotope is 177Hf.

Beta emission of a radioisotope yields a  daughter nuclide whose amass number is the same as that of its parent nucleus but its atomic number is greater is greater than that of the parent nucleus by 1 unit.

Also, gamma emission does not lead to any change in the mass number of atomic number of the daughter nucleus produced.

Hence, the stable daughter nuclide, 177Hf is produced.

Learn more: brainly.com/question/1770619

4 0
3 years ago
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

4 0
3 years ago
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