Compounds are not possible to be separated by physical means
<u>A. reaction A: chemical; reaction B: nuclear</u>
Answer:
Solution given:
as
Circle 2: Reaction B
-Involves neutrons
-Happens inside atomic nucleus
-Releases relatively large amounts of energy
<u>I</u><u>t</u><u> </u><u>h</u><u>a</u><u>p</u><u>p</u><u>e</u><u>n</u><u>s</u><u> </u><u>o</u><u>n</u><u> </u><u>n</u><u>o</u><u>r</u><u>m</u><u>a</u><u>l</u><u> </u><u>c</u><u>h</u><u>e</u><u>m</u><u>i</u><u>c</u><u>a</u><u>l</u><u> </u><u>reaction</u><u>.</u>
Circle 2: Reaction B
-Involves neutrons
-Happens inside atomic nucleus
-Releases relatively large amounts of energy
<u>I</u><u>t</u><u> </u><u>h</u><u>a</u><u>p</u><u>p</u><u>e</u><u>n</u><u>s</u><u> </u><u>o</u><u>n</u><u> </u><u>n</u><u>u</u><u>c</u><u>l</u><u>e</u><u>a</u><u>r</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>i</u><u>o</u><u>n</u><u>.</u>
Mixtures and solutions... what?
C=λν ==> λ=c/ν
λ= 3x10^8 m/s / 1.50 x 10^15 s^-1
λ= 2 x 10^-7 m
λ= 200 nm
Answer:
Mass of paraffin that would be needed to collect the amount of energy as 4.73 x 10³ kg of water is 6.85 * 10³ kg
Explanation:
Specific heat capacity of paraffin, Cp = 2.90 J/g.°C = 2900 J/kg.°C
Specific heat capacity of water = 4.20 J/g.°C = 4200 J/kg.°C
Amount of heat energy that can be stored by 4.73 * 10³ kg of water heated through a degree rise in temperature can be calculated as below:
Using H = mCpθ where m is mass of substance, Cp is specific heat capacity and θ is temperature change
H = 4.73 * 10³ kg * 4200 J/kg.°C * 1°C
H = 19866000 J
Mass of paraffin that would be needed to collect the amount of energy as 4.73 x 10 3 kg of water for a degree rise in temperature is calculated as follows:
H = mCpθ
19866000J = m * 2900 J/kg.°C * 1°C
m = 19866000J / 2900 J/kg
m = 6850.34 kg = 6.85 * 10³ kg
Therefore, mass of paraffin that would be needed to collect the amount of energy as 4.73 x 10³ kg of water is 6.85 * 10³ kg