C. Single-replacement
Chlorine replaces Bromine in KBr.
Answer:
Abundance of 32S is 94.41%
Explanation:
The average atomic mass is defined as the sum of the atomic masses of each isotope times its abundance:
Average atomic mass = ∑ Atomic mass istope*Abundance
For the sulfur:
32.07amu = 31.97207X + 32.97146Y + 33.96786*0.0422 <em>(1)</em>
<em>Where X is abundance of 32S and Y abundance of 33S</em>
Also we can write:
1 = X + Y + 0.0422 <em>(2)</em>
0.9578 - X = Y
Because the sum of the abundances = 1
Replacing (2) in (1):
32.07amu = 31.97207X + 32.97146(0.9578 - X) + 33.96786*0.0422
32.07 = 31.97207X + 31.58006 - 32.97146X + 1.43344
-0.9435 = -0.99939X
0.9441 =X
In percentage, abundance of 32S is 94.41%
The equilibrium constant Kc for this reaction is calculated as follows
from the equation N2 + 3H2 =2 NH3
qc = (NH3)2/{(N2)(H2)^3}
Qc is therefore = ( 0.001)2 /{(0.1) (0.05)^3} = 0.08
Answer: 3.69 × 10^27
Explanation:
Amount of energy required = 7.06 × 10^4 J
Frequency of microwave (f) = 2.88 × 10^10 s−1
Planck's constant (h) = 6.63 × 10^-34 Jᐧs/quantum
Recall ;
Energy of photon = hf
Therefore, energy of photon :
(6.63 × 10^-34)j.s× (2.88 × 10^10)s^-1
= 19.0944 × 10^(-34 + 10) = 19.0944×10^-24 J
Hence, number of quanta required :
(7.06 × 10^4)J / (19.0944 × 10^-24)J
= 0.369 × 10^(4 + 24) = 0.369×10^28
= 3.69 × 10^27
<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>