Answer: V2= 15.0403226 Liters
Explanation:
Use V1/T1=V2/T2
Make sure you change the degrees Celsius to Kelvin. (Kelvin = degrees Celsius +273)
10.0L / 248 K = V2/ 373 K
Cross multiply V1 and T2 and divide by T1
(10.0 L)( 373K)/ 248 K = V2
V2= 15.0403226 Liters (Kelvin cancels out)
Answer:
Of bond is formed by sharing of electrons - covalent
If bond is formed by donation and by accepting electrons - ionic bond
Bond formed between metals - metallic bond.
The answer would be .5 mols because you take the total amount of grams, which is 20, and you had up the molar mass of sodium hydroxide, which would be 40. After you have this you would set this up as a stochiometry equation. With 1 mol on top you dived 20/40 to cancel out your grams. This leaves you with .5 mols
Answer:
a) the final kilocalories per gram for food will be less because the mass was reduced
b)the final kilocalories per gram for food will be less since
c) the final kilocalories per gram for food will be less since the reaction will eventually go to completion
d) the final kilocalories per gram for food will be more.
Explanation:
a) the final kilocalories per gram for food will be less because the mass was reduced from 110.3 to 101.3g
b)the final kilocalories per gram for food will be less since some marshmallow fell off before the reaction
c) the final kilocalories per gram for food will be less since the reaction will eventually go to completion
d) the final kilocalories per gram for food will be more since the thermometer that got stuck will add to the value of final kilocalories per gram
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>