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
#learnwithBrainly
Answer:
When our bodies are dry and wind blows by, we lose some energy to the air molecules. When are bodies are wet, we have a substance on our skin that likes to absorb heat. So when wind blows by, we lose a LOT of energy to the air molecules. When the body loses heat energy, our body temperature drops.
Explanation:
hope it helps
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175.8 g NaCl to moles:
(175.8 g)/(58.44 g/mol) = 3.008 mol NaCl
Molarity = moles of solute/volume of solution in liters
(3.008 mol NaCl)/(1.5 L) = 2.0 M.
The molarity of this solution would be 2.0 M.
Answer:
60.0mL
Explanation:
For the process of titration , the formula we use is -
M₁V₁ = M₂V₂
Where ,
M₁ = initial concentration
V₁ = initial volume
M₂ = final concentration
V₂ = final volume .
Hence , from the question ,
The given data is ,
M₁ = 0.050 M
V₁ = 30.0mL
M₂ = 0.025 M
V₂ = ?
Now, to determine the unknown quantity , the formula can be applied ,
Hence ,
M₁V₁ = M₂V₂
Putting the respective values ,
0.050 M * 30.0mL = 0.025 M * ?
solving the above equation ,
V₂ = ? = 60.0mL