Answer:
THE HEAT REQUIRED TO CHANGE 2 KG OF ETHANOL FROM 26 °C TO THE BOILING POINT AND TO VAPOR AT THAT TEMPERATURE IS 1965.175 KJ.
Explanation:
Boiling point of ethanol = 78.29 °C = 78.29 + 273 K = 351.29 K
Mass = 2 kg = 2000 g
Final temp. = 26.0 °C = 26 + 273 K= 299 K
Change in temperature = (78.29 - 26) °C = 52.29 °C
1. Heat required to raise the temperature from 26 °C to the boiling point?
Heat = mass * specific heat * change in temperature
Heat = 2000 * 2.44 * 52.29
Heat = 255 175.2 J
2. Heat required to change the liquid to vapor at that temperature?
Heat = mass * enthalphy of vaporization
Heat = 2000 * 855
Heat =1 710000 J
The total heat required to raise the temperature of 2 kg of ethanol from 26 °C to the boiling point and then to change the liquid to vapor at that temperature will be:
Heat = mcT + m Lv
Heat = 255 175.2 J + 1710000 J
Heat = 1965175.2 J
Heat = 1965.175 kJ of heat.
This may seem confusing because they give you two masses, but all you have to do is pick one to do the calculations. Personally, I would pick O2, since the molar mass is easier to calculate. The answer would be 3.3 g (rounded for sig figs). To get this, first take the 5.9 grams of O2 and convert it to moles by dividing by the molar mass of oxygen gas, which is 32. Then, multiply both by the mole-mole ratio, which is 2:2, or simply 1:1. After that, multiply that by 18g, which is the molar mass of water to get grams of water.
REMEMBER, you have to write and balance the chemical equation before you can do any of that work.
That happens to be CH4 + 2O2 => CO2 + 2H2O
Answer:
The correct answer is 2.016 x 10⁻¹⁷
Explanation:
We have the following chemical reactions and their equilibrium constants (K):
(1) H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + HCO₃⁻(aq) K₁= 4.20×10⁻⁷
(2) HCO₃⁻(aq) + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
And we have to obtain K for the following reaction:
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq)
If we add equations (1) and (2) we obtain the the desired equation. Remember that when we add chemical equations, the global equilibrium constant is the product of the constants.
H₂CO₃(aq) + H₂O ⇒ H₃O⁺(aq) + <em><u>HCO₃⁻(aq)</u></em> K₁= 4.20×10⁻⁷
+
<em><u>HCO₃⁻(aq)</u></em> + H₂O ⇒ H₃O⁺(aq) + CO₃²⁻(aq) K₂= 4.80×10⁻¹¹
-------------------------------------------------------------
H₂CO₃(aq) + 2H₂O ⇒ 2H₃O⁺(aq) + CO₃²⁻(aq) K= K₁ x K₂
K = K₁ x K₂ = (4.20×10⁻⁷) x (4.80×10⁻¹¹) = 2.016 x 10⁻¹⁷
The answer is TiBr4 which is the formula
As size increases<span> or decreases, </span>DENSITY REMAINS THE SAME<span>! As a rule, things that are MORE dense will sink or fall to the bottom of things that are LESS dense (think of oil and water). ... Notice that </span>if<span> you DECREASE the </span>mass<span> of something or</span>INCREASE<span> the </span>volume<span>, the </span>density<span> will be less…</span>
