Answer:
Part A
Kp = 3.4 x 10⁴
Part B
Kp = 2.4 x 10⁻¹⁴
Part C
Kp = 1.2 x 10⁹
Explanation:
2PH₃(g) + As₂(g) ⇌ 2 AsH₃(g) + P₂(g) Kp = 2.9 x 10⁻⁵
Kp = [AsH₃]²[P₂]/[PH₃]²[As] = 2.9 x 10⁻⁵
Part A
it is the inverse of the equilibrium given
Kp(A) = 1/ Kp = 1 / 2.9 x 10⁻⁵ = 3.4 x 10⁴
Part B
Is the equilibrium where the coefficients have been multiplied by 3,
Kp(B) = ( Kp )³ = ( 2.9 x 10⁻⁵ )³ = 2.4 x 10⁻¹⁴
Part C
This is the reverse equilibrium multipled by 2.
Kp(C) = ( 1/Kp)² = ( 1/ 2.9 x 10⁻⁵ )² = 1.2 x 10⁹
Answer:
NH₄CN, NH₄IO₃, Fe(CN)₃, Fe(IO₃)₃
Explanation:
Cations (positively charged ions) can only form ionic bonds with anions (negatively charged ions). However, you can't just simply put one cation and one anion together to form a compound. Each compound needs to been neutral, or have an overall charge of 0. When cations and anions do not have charges that perfectly cancel, you need to modify the amount of each ion in the compound.
1.) NH₄CN
-----> NH₄⁺ and CN⁻
-----> +1 + (-1) = 0
2.) NH₄IO₃
-----> NH₄⁺ and IO₃⁻
-----> +1 + (-1) = 0
3.) Fe(CN)₃
-----> Fe³⁺ and CN⁻
-----> +3 + (-1) + (-1) + (-1) = 0
4.) Fe(IO₃)₃
-----> Fe³⁺ and IO₃⁻
-----> +3 + (-1) + (-1) + (-1) = 0
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
<span>Co-product (rather than side product) of Horner-Emmons Wittig reaction is alpha-hydroxy phosphonate which by definition is hydrophilic as opposed to triphenylphosphine oxide. In addition, the side groups in Horner-Wittig reagents are typically Me or Et which are less bulky than phenyls are more compatible with hydrophilicity.</span>
