<h3>
<u>ANSWER</u></h3>
2. Neon, only
<h3>
<u>EXPLANATION</u></h3>
When Na and F combine they form an electron configuration of 2-8. Na electron configuration is 2-8-1 while F is 2-7, so when they form an ionic bond F will gain Na outermost electron to complete its valence shell due to having a higher negativity. Neon has an electron configuration of 2-8 but argon has 2-8-8.
the volume will be 0.84mol of Y
Answer:
25 mM Tris HCl and 0.1% w/v SDS
Explanation:
A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:
(250 mM Tris HCl)/10 = 25 mM Tris HCl
(1.92 M glycine)/10 = 0.192 M glycine
(1% w/v SDS)/10 = 0.1% w/v SDS
Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.
To find the molecular formula of the compound, we get the molar mass of empirical formula CH2o which is 30 g/mol. We divided the molecular mass of 90.087 g/mol by 30, we get 3. Hence, we multiply each element by 3, getting the molecular formula of C3H6O3.
Answer:
Moles NH₃: 0.0593
0.104 moles of N₂ remain
Final pressure: 0.163atm
Explanation:
The reaction of nitrogen with hydrogen to produce ammonia is:
N₂ + 3 H₂ → 2 NH₃
Using PV = nRT, moles of N₂ and H₂ are:
N₂: 1atmₓ3.0L / 0.082atmL/molKₓ273K = 0.134 moles of N₂
H₂: 1atmₓ2.0L / 0.082atmL/molKₓ273K = 0.089 moles of H₂
The complete reaction of N₂ requires:
0.134 moles of N₂ × (3 moles H₂ / 1 mole N₂) = <em>0.402 moles H₂</em>
That means limiting reactant is H₂. And moles of NH₃ produced are:
0.089 moles of H₂ × (2 moles NH₃ / 3 mole H₂) = <em>0.0593 moles NH₃</em>
Moles of N₂ remain are:
0.134 moles of N₂ - (0.089 moles of H₂ × (1 moles N₂ / 3 mole H₂)) = <em>0.104 moles of N₂</em>
And final pressure is:
P = nRT / V
P = (0.104mol + 0.0593mol)×0.082atmL/molK×273K / 5.0L
<em>P = 0.163atm</em>
