Answer is: chemical elements<span> in </span>group 15<span> of the </span>periodic table have <span>neutral atom with a valence electron configuration of ns</span>²<span>np</span>³<span>.
</span>Nitrogen: [He] 2s² 2p³.
Phosphorus: [Ne] 3s² 3p³.
Arsenic: <span>[Ar] 3d</span>¹⁰ 4s² 4p³.
Bismuth: <span>[Xe] 4f</span>¹⁴ 5d¹⁰ 6s² 6p³.
Moscovium: predicted [Rn] 5f¹⁴<span> 6d</span>¹⁰<span> 7s</span>² 7p³.
These elements have 5 valence electrons.
Answer:
Average atomic mass = 15.86 amu.
Explanation:
Given data:
Number of atoms of Z-16.000 amu = 205
Number of atoms of Z-14.000 amu = 15
Average atomic mass = ?
Solution:
Total number of atoms = 205 + 15 = 220
Percentage of Z-16.000 = 205/220 ×100 = 93.18%
Percentage of Z-14.000 = 15/220 ×100 = 6.82 %
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (93.18×16.000)+(6.82×14.000) /100
Average atomic mass = 1490.88 + 95.48 / 100
Average atomic mass = 1586.36 / 100
Average atomic mass = 15.86 amu.
Answer:
38.75 L
Explanation:
From the question,
Applying Boyles Law,
PV = P'V'....................... Equation 1
Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' = Final Volume of Argon gas.
make V the subject of the equation
V = P'V'/P.................... Equation 2
Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.
Substitute these values into equation 2
V = (456×2.94)/34.6
V = 38.75 L
1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol
</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO
<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.
CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)
How much
we need CO? 3 mol 6 mol
We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.
3) </span>CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol
4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol
2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
Answer:
68.1% is percent yield of the reaction
Explanation:
The reaction of methane with oxygen is:
CH₄ + 2O₂ → CO₂ + 2H₂O
<em>Where 2 moles of oxygen react per mole of CH₄</em>
<em />
Percent yield is:
Actual yield (28.2g CO₂) / Theoretical yield * 100
To solve this question we need to find theoretical yield finding limiting reactant :
<em>Moles CH₄:</em>
15.1g CH₄ * (1mol / 16.04g) = 0.9414 moles
<em>Moles O₂:</em>
81.2g * (1mol / 32g) = 2.54 moles
For a complete reaction of 0.9414 moles of CH₄ are needed:
0.9414 moles CH₄ * (2 mol O₂ / 1mol CH₄) = 1.88 moles of O₂. As there are 2.54 moles, O₂ is in excess and <em>CH₄ is limiting reactant</em>
In theoretical yield, the moles of methane added = Moles of CO₂ produced. That is 0.9414 moles CO₂. In grams = Theoretical yield:
0.9414 moles CO₂ * (44.01g / mol) = 41.43g CO₂
Percent yield: 28.2g CO₂ / 41.43g CO₂ * 100=
<h3>68.1% is percent yield of the reaction</h3>
