1 mole of H2O weighs 18 g
therefore 13.8 g of liquid H2O = 13.8/18 moles
ΔHvaporization = 40.65 kJ/mol
heat required to change 13.8 grams of H2O from a liquid to a gas at 100 degrees Celsius = 40.65 x 13.8/18 = 31.165 kJ
Answer:
Bismuth, because when three cubes of the same mass were heated and dropped into the same temperature water, the water the bismuth cube was dropped in heated up much less than the water with iron or tin.
REWRITE THIS IN YOUR OWN WORDS PLEASE, THANK YOU.
Credits to: www.generationgenius.com › wp-content › uploads
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2M was neutralized by 0.01 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 0.125 mL
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl (Stomach acid)
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 0.125 mL
Answer:
612 K
Explanation:
From the question given above, the following data were obtained:
Initial temperature (T₁) = 306 K
Initial pressure (P₁) = 150 kPa
Final pressure (P₂) = 300 kPa
Volume = 4 L = constant
Final temperature (T₂) =?
Since the volume is constant, the final (i.e the new) temperature of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
150 / 306 = 300 / T₂
Cross multiply
150 × T₂ = 306 × 300
150 × T₂ = 91800
Divide both side by 150
T₂ = 91800 / 150
T₂ = 612 K
Thus, the new temperature of the gas is 612 K
Explanation:
hope for help please braimliesy
