Answer:
-3.617 °C
Explanation:
Step 1: Given data
Mass of water (m): 210.0 g
Energy released in the form of heat (Q): -3178 J (the minus sign corresponds to energy being released)
Specific heat of water (c): 4.184 J/g.°C
Temperature change (ΔT): ?
Step 2: Calculate the temperature change
We will use the following expression.
Q = c × m × ΔT
-3178 J = 4.184 J/g.°C × 210.0 g × ΔT
ΔT = -3.617 °C
55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
Answer:
A. The balloons will increase to twice their original volume.
Explanation:
Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:
P ∝ 1/V
P = k/V
PV = k (constant)
P = pressure, V = volume.
Let the initial pressure of the balloon be P, i.e. , initial volume be V, i.e. . The pressure is then halved, i.e.
Therefore the balloon volume will increase to twice their original volume.
Answer: 0.151
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Given:
Putting in the values we get:
Thus the rate of appearance of is 0.151