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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Answer:
I believe it is "Arsenenate"
Answer: 162.8 grams
Explanation:
Magnesium nitrate has a chemical formula of Mg(NO3)2.
Given that:
Number of moles of Mg(NO3)2 = 1.1 moles
Mass in grams of Mg(NO3)2 = ?
For Molar mass of Mg(NO3)2, use atomic mass of magnesium = 24g, nitrogen = 14g, oxygen = 16g
Mg(NO3)2 = 24g + (14g + 16gx3) x 2
= 24g + (14g + 48g) x 2
= 24g + (62g) x 2
= 24g + 124g
= 148g/mol
Now, apply the formula:
Number of moles = Mass in grams / molar mass
1.1 moles = Mass / 148g/mol
Mass = 1.1 moles x 148g/mol
Mass = 162.8 grams
Thus, there are 162.8 grams of magnesium nitrate.