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>
Answer:
90g of H2O
Explanation:
2H2 + O2 —> 2H2O
First, we calculate the molar masses of H2 And H20.
Molar Mass of H2 = 2g/mol
Mass conc of H2 from the balanced equation = 2 x 2 = 4g
Molar Mass of H2O = 2 + 16 = 18g/mol
Mass conc of H2O from the balanced equation = 2x18 = 36g
From the equation,
4g of H2 produced 36g of H2O
Therefore, 10g of H2 will be produce = (10x36)/4 = 90g of H2O
The reaction of baking soda or baking powder with the liquid in the batter: These ingredients react together and cause air bubbles to form. ... Heat of the oven: The heat of the oven can cause baking powder to react further and cause more air bubbles, and the heat also sets the structure of the cake.