Answer:
Ka = 6.02x10⁻⁶
Explanation:
The equilibrium that takes place is:
We <u>calculate [H⁺] from the pH</u>:
- [H⁺] =

Keep in mind that [H⁺]=[A⁻].
As for [HA], we know the acid is 0.66% dissociated, in other words:
We <u>calculate [HA]</u>:
Finally we <u>calculate the Ka</u>:
- Ka =
= 6.02x10⁻⁶
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of
) =-1275.0
enthalpy of combustion of oxygen(Δ
of
) = zero
enthalpy of combustion of carbon dioxide(Δ
of
) = -393.5
enthalpy of combustion of water(Δ
of
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ
= ∈Δ
(products) - ∈Δ
(reactants)
(s) +6
(g) → 6
(g)+ 6
(l)
Δ
= [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ
= [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ
= 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ
= -2361 - 1714 - 0 + 1275
Δ
=-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer:
m = 180 g
Explanation:
Given data:
Energy absorbed = 108 J
Mas of gold = ?
Initial temperature = 25°C
Final temperature = 29.7 °C
Specific heat capacity of gold = 0.128 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT =29.7 °C - 25°C
ΔT = 4.7 °C
108 J = m ×0.128 J/g.°C ×4.7 °C
108 J = m ×0.60 J/g
m = 108 J/0.60 J/g
m = 180 g
Answer:
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Explanation:
As molecules heat they begin to move faster. The heat from the metal plate will make the molecules at room temperature move faster. And the room temperature makes the hot place cool, making the hot plate molecules slow down.
To visualize this, you can use this link
https://phet.colorado.edu/en/simulations/gas-properties
click the play button to activate the activity