If a<span> student used 10 mL water instead of 30 mL for extraction of salt water from mixture, the extraction of salt will be lesser than compared to using 30 mL since less solute will dissolve in 10 mL.</span>
Answer:
5 seconds
Explanation:
Speed x Time. So t=ds. t=51=5.
20600Cal
Explanation:
Given parameters:
Mass of water = 319.5g
Initial temperature = 35.7°C
Final temperature = 100°C
Unknown:
Calories needed to heat the water = ?
Solution:
The calories is the amount of heat added to the water. This can be determined using;
H = m c Ф
c = specific heat capacity of water = 4.186J/g°C
H is the amount of heat
Ф is the change in temperature
H = m c (Ф₂ - Ф₁)
H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J
Now;
1kilocalorie = 4184J
85996.56J to kCal;
= 20.6kCal = 20600Cal
learn more:
Specific heat brainly.com/question/3032746
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The heat transfer formula is;
Q = m * c * Δ T >>>> (1)
where, Q is the heat transfer
m = mass (gram)
c = the specific heat capacity (J/g)
Δ T = change in temperature
∵ we have one mole of Ethanol
∴ the weight of ethanol equals its molecular weight = (2*12)+(6*1)+(16) = 46 g
we will assume that the specific heat capacity of ethanol is 2.46 J/g (from google)
ΔT = 25 - 320 = - 295 C
By substitution in (1)
∴ Q = 2.46 * 46 * (-295) = - 33382.2 J