A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.
given that :
8 ml of a 5 % solution mix with 10 ml . that means the 80 mL of 5 % solution is diluted with water of 10 mL
therefore, 80 × 5 = 10 × x %
x % = 40 %
Therefore, the final percentage strength of the solution is 40 %
Thus, A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.
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Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.
Answer:
Q = 60192 j
Explanation:
Given data:
Volume of water = 0.45 L
Initial temperature = 23°C
Final temperature = 55°C
Amount of heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 55°C - 23°C
ΔT = 32°C
one L = 1000 g
0.45 × 1000 = 450 g
Specific heat capacity of water is 4.18 j/g°C
Q = m.c. ΔT
Q = 450 g. 4.18 j/g°C. 32°C
Q = 60192 j
As the gas cools it condenses and becomes a liquid its atoms also become smaller
