Answer:
189 Joules
Explanation:
Applying,
Q = cm(t₂-t₁)............. equation 1
Where Q = Heat, c = specific heat capacity of water, m = mass of water, t₁ = Initial Temperature, t₂ = Final temperature.
From the question,
Given: m = 15 grams = 0.015 kg, t₁ = 21 °C, t₂ = 24 °C
Constant: c = 4200J/kg.°C
Substitute these values into equation 1
Q = 0.015×4200×(24-21)
Q = 0.015×4200×3
Q = 189 Joules
You can predict it based of the electronegativity
Answer: The new pressure of the gas in Pa is 388462
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas at STP = 
= final pressure of gas = ?
= initial volume of gas = 700.0 ml
= final volume of gas = 200.0 ml
= initial temperature of gas = 273 K
= final temperature of gas = 
Now put all the given values in the above equation, we get:
The new pressure of the gas in Pa is 388462
Answer is <span>C: It's appearance changed.</span>
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>
