Answer:
negative
Explanation:
When something slows down, its acceleration is the opposite of the velocity.
Answer:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released or absorbed by the system.
m is the mass of the ice (m = 60.0 g).
c is the specific heat capacity of ice (c = 2.108 J/g.°C).
ΔT is the temperature difference (ΔT = 1.0 °C).
∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.
<em>It is known that 1.0 cal = 4.18 J.</em>
<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>
1 mols of Aluminium ion forms 1 mol aluminium phosphate
Molar mass of AlPO_4
Moles of AlPO_4
- 61µg/106
- 0.000061/106
- 5.75×10^{-7}
- 57.5µmol
Moles of Al3+=57.5µmol
Answer:
29260J
Explanation:
Given parameters:
Mass of water sample = 100g
Initial temperature = 30°C
Final temperature = 100°C
Unknown:
Energy required for the temperature change = ?
Solution:
The amount of heat required for this temperature change can be derived from the expression below;
H = m c (ΔT)
H is the amount of heat energy
m is the mass
c is the specific heat capacity of water = 4.18J/g°C
ΔT is the change in temperature
Now insert the parameters and solve;
H = 100 x 4.18 x (100 - 30)
H = 100 x 4.18 x 70 = 29260J
