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goldenfox [79]
3 years ago
9

Convert 12.8 cm to km.

Chemistry
2 answers:
Makovka662 [10]3 years ago
7 0

Answer:

0.000128 kilometers

Andre45 [30]3 years ago
7 0

Answer:

0.000128 kilometre..

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In the reaction shown above , the?
matrenka [14]

Answer:

I think option (d) is right answer

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3 years ago
A tau lepton decays into an electron, an electron antineutrino and a tau neutrino. Write out this reaction in symbolic (equation
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Explanation:

hhidhwvagaue hrurifv3v2vd

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3 years ago
If a patient has a medical condition that causes his cells to absorb fewer than normal molecules, this patient would likely feel
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Answer:oxygen Explanation:The medical condition described here is anaemia. It is a blood cell disorder whereby the red blood cell doesn't function properly and hence doesn't carry enough oxygen to the tissues. This is usually caused when ones body is deficient of iron.The symptoms that may occur to such patients are weakness, fatigue, headache and pale skin.Based on the explanation, the answer is oxygen

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2 years ago
Balance the equation <br> NaOH + CH3COOH
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NaOH + CH3COOH -> CH3COONa + H20

8 0
4 years ago
An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder
seraphim [82]

Answer:

\Delta H=-11897J

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

\Delta H=\Delta U+V\Delta P

Whereas the change in the internal energy is computed by:

\Delta U=nCv\Delta T

So we compute the initial and final temperatures for one mole of the ideal gas:

T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K  }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K  }{n}

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J

Then, the volume-pressure product in Joules:

V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J

Finally, the change in the enthalpy for the process:

\Delta H=-7140J-4757J\\\\\Delta H=-11897J

Best regards.

7 0
4 years ago
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