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KatRina [158]
4 years ago
6

Tearing paper into pieces is an example of what kind of change. Please help

Chemistry
1 answer:
FrozenT [24]4 years ago
8 0
This is a physical change.
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What energy resource can be used instead of fossil fuel
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We could use solar power, wind power, geothermal power, hydroelectric power, or nuclear power. There are probably more but this is what I can think of off the top of my head. I hope this helps. Let me know if anything is unclear.
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3 years ago
The colour of RBC is Red .why?​
RoseWind [281]

Answer: fat show the first half and a lot of people in a row in this rr

7 0
3 years ago
Express 749 000 000 in scientific notation,
Dima020 [189]

Answer:

7.49 × 108

Explanation:

Scientific notation is a way to express numbers in a form that makes numbers that are too small or too large more convenient to write. It is commonly used in mathematics, engineering, and science, as it can help simplify arithmetic operations. In scientific notation, numbers are written as a base, b, referred to as the significant, multiplied by 10 raised to an integer exponent, n, which is referred to as the order of magnitude:

8 0
3 years ago
For the gas phase decomposition of 1-bromopropane, CH3CH2CH2BrCH3CH=CH2 + HBr the rate constant at 622 K is 6.43×10-4 /s and the
ladessa [460]

<span>Answer is: activation energy of this reaction is 212,01975 kJ/mol.
Arrhenius equation: ln(k</span>₁/k₂) = Ea/R (1/T₂ - 1/T₁<span>).
k</span>₁<span> = 0,000643 1/s.
k</span>₂ = 0,00828 1/s.

T₁ = 622 K.

T₂ = 666 K.

R = 8,3145 J/Kmol.

1/T₁<span> = 1/622 K = 0,0016 1/K.
1/T</span>₂<span> = 1/666 K = 0,0015 1/K.
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol · (-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol · (-0,0001 1/K).
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>

4 0
3 years ago
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1·M−1s−1 : →2SO3g+2SO2gO2g Suppo
RoseWind [281]

Answer:

[SO_3]=0.25M

Explanation:

Hello there!

In this case, since the integrated rate law for a second-order reaction is:

[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}

Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:

[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\

[SO_3]=0.25M

Best regards!

4 0
3 years ago
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