Answer:
Ammonium bisulfide, NH4HS , forms ammonia, NH3 , and hydrogen sulfide, H2S , through the reaction NH4HS(s)⇄NH3(g)+H2S(g) This reaction has a Kp value of 0.120 at 25°C .
An empty 5.00-L flask is charged with 0.300 g of pure H2S(g) , at 25°C
Partial pressure of NH3 = 0.325 atm
Partial pressure of H2S = 0.368 atm
Explanation:
Temperature = 25 °C = 298 K
Volume = 5 L
Mass of H2S = 0.30 g
Moles of H2S = 0.30 / 34.08
= 0.0088
Using PV = nRT
Initial pressure of H2S = 0.0088 * 0.0821 * 298 / 5
= 0.043 atm
Kp = 0.120
(a). NH4HS(s) ⇄ NH3(g) + H2S(g)
Initial - 0.0 0.043
Change - + x + x
Equilibrium - x 0.043 + x
Kp = P(NH3) * P(H2S)
0.120 = x (0.043 + x)
x = 0.325 atm
Hence, at equilibrium:
Partial pressure of NH3 = 0.325 atm
Partial pressure of H2S = 0.043 + 0.325 = 0.368 atm
<em><u>Larger molecules take longer to move up the chromatography paper or TLC plate, whereas smaller molecules are more mobile. Likewise, the polarity of the molecules can affect how far the spots travel, depending on the type of solvent used.</u></em>
<span>Sodium carbonate (Na2CO3) reacts with acetic acid (CH3COOH) to form sodium acetate (NaCH3COO), carbon dioxide (CO2), and water (H2O). A chemist carries out this reaction in a bomb calorimeter. The reaction causes the temperature of a bomb calorimeter to decrease by 0.985 K. The calorimeter has a mass of 1.500 kg and a specific heat of 2.52 J/g K. What is the heat of reaction for this system? What equation should I use in this case? I've written down these notes: Steps: 1. Calculate the mass of the solution in total. 2. Convert mass to volume or vice versa if needed. 3. Calculate the temperature change of the solution. 4. Calculate the energy released by the reaction.</span>
Answer:
A.) Brønsted-Lowry bases
Explanation:
Amines have a lone pair of electrons.
Brønsted-Lowry bases donate a lone pair of electrons in exchange for a hydrogen ion.
Therefore, if exposed to an acid, amines will give up electrons in order to bond with a hydrogen. This makes them Brønsted-Lowry bases.
Answer:
Top left: RR Top right: Rr Bottom left: Rr Bottom right: rr
Explanation:
if both parents are heterozygous then the table will always look like this. the chances of the offspring being heterozygous are 50%. the offspring has a 25% chance of having two recessive traits and a 25% chance of having two dominant traits.
