Answer:
D. 15g
Explanation:
The law of conservation of mass states that, in a chemical reaction, mass can neither be created nor destroyed. This means that the amount of matter in the elements of the reactants must be equal to the amount in the resulting products.
In this question, 25 grams of a reactant AB, was broken down in a reaction to produce 10 grams of products A and X grams of product B. According to the law of conservation of mass, the mass of the reactant must be equal to the total mass of the products. This means that 25 grams must also be the total mass of both products in this reaction. Hence, if product A is 10 grams, product B will be 25 grams - 10 grams = 15 grams.
Therefore, product B must be 15 grams in order to form a total of 25 grams when added to the mass of product A. This will equate the mass of the reactant AB and fulfill the law of conservation of mass.
Answer:
The P-H bonds are more polar than the N-H bonds.
Explanation:
Phosphine is a polar molecule with non-polar bonds. The phosphorus atom is bonded to three hydrogen atoms and the phosphorus atom has a lone pair of electrons. Since hydrogen and phosphorus are equal in electronegativity, it implies that they attract the shared pairs of electrons the same amount,hence bonding electrons are shared equally making the covalent bonds non-polar.
The lone pair of electrons on phosphorus causes the molecule to be asymmetrical with respect to charge distribution this is why the molecule is polar even though the are non-polar bonds in the molecule.
Looking at the values of electro negativity stated in the question, one can easily see that the difference in electro negativity between nitrogen and hydrogen is 0.9 while the difference in electro negativity between phosphorus and hydrogen is zero. It is clear that NH3 is naturally more polar than PH3 since each individual N-H bond in NH3 is a polar bond while the individual P-H bonds in PH3 are nonpolar.
Explanation:
It is known that efficiency is denoted by
.
The given data is as follows.
= 0.82,
= (21 + 273) K = 294 K
= 200 kPa,
= 1000 kPa
Therefore, calculate the final temperature as follows.
0.82 =
= 1633 K
Final temperature in degree celsius =
= 
Now, we will calculate the entropy as follows.

For 1 mole, 
It is known that for
the value of
= 0.028 kJ/mol.
Therefore, putting the given values into the above formula as follows.

= 
= 0.0346 kJ/mol
or, = 34.6 J/mol (as 1 kJ = 1000 J)
Therefore, entropy change of ammonia is 34.6 J/mol.
Answer:
i. n = 5
ii. ΔE = 7.61 ×
KJ/mole
Explanation:
1. ΔE = (1/λ) = -2.178 ×
(
-
)
(1/434 ×
) = -2.178 ×
(
)
⇒ 434 ×
= (1/-2.178 ×
)
But,
= 2
434 ×
= (1/2.178 ×
)
434 ×
× 2.178 ×
= 
⇒
= 5
Therefore, the initial energy level where transition occurred is from 5.
2. ΔE = hf
= (hc) ÷ λ
= (6.626 × 10−34 × 3.0 ×
) ÷ (434 ×
)
= (1.9878 ×
) ÷ (434 ×
)
= 4.58 ×
J
= 4.58 ×
KJ
But 1 mole = 6.02×
, then;
energy in KJ/mole = (4.58 ×
KJ) ÷ (6.02×
)
= 7.61 ×
KJ/mole
Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M