To get the theoretical yield of ammonia NH3:
first, we should have the balanced equation of the reaction:
3H2(g) + N2(g) → 2NH3(g)
Second, we start to convert mass to moles
moles of N2 = N2 mass / N2 molar mass
= 200 / 28 = 7.14 moles
third, we start to compare the molar ratio from the balanced equation between N2 & NH3 we will find that N2: NH3 = 1:2 so when we use every mole of N2 we will get 2 times of that mole of NH3 so,
moles of NH3 = 7.14 * 2 = 14.28 moles
finally, we convert the moles of NH3 to mass again to get the mass of ammonia:
mass of NH3 = no.moles * molar mass of ammonia
= 14.28 * 17 = 242.76 g
T = 20 % : 20 / 100 = 0.2
m1 = solute
m2 = Solvent
T = m1 / m1 + m2
0.2 = 500 g / 500 g + m2
0.2 * ( 500 + m2 ) = 500
0.2 * 500 + 0.2 m2 = 500
100 + 0.2 m2 = 500
0.2 m2 = 500 - 100
0.2 m2 = 400
m2 = 400 / 0.2
m2 = 2000 g of water
hope this helps!
Answer:
b) 3.10
Explanation:
HF ⇄ H
+ + F
Using Henderson-Hasselbalch Equation:
pH = pKa + log [A-]/[HA].
Where;
pKa = Dissociation constant = -log Ka
Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266
[A-] = concentration of conjugate base after dissociation = moles of base/total volume
= 0.15 x 0.3/0.8
= 0.05625 M
[HA] = concentration of the acid = moles of acid/total volume
= 0.10 x 0.5/0.8
= 0.0625 M
Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>
pH = 3.14266 + log [0.05625/0.0625]
= 3.14267 + (-0.04575749056)
= 3.09691250944
<em>From all the available options below:</em>
<em>a) 2.97
</em>
<em>b) 3.10
</em>
<em>c) 3.19
</em>
<em>d) 3.22
</em>
<em>e) 3.32</em>
The correct option is b.
Answer:
See the explanation
Explanation:
In this case, in order to get an <u>elimination reaction</u> we need to have a <u>strong base</u>. In this case, the base is the phenoxide ion produced the phenol (see figure 1).
Due to the resonance, we will have a more stable anion therefore we will have a less strong base because the negative charge is moving around the molecule (see figure 2).
Finally, the phenoxide will attack the <u>primary carbon</u> attached to the Cl. The C-Cl bond would be broken and the C-O would be produced <u>at the same time</u> to get a substitution (see figure 1).
When 6.85×10⁵ cal is converted to kilojoules, the result obtained is 2866.04 KJ
<h3>Data obtained from the question </h3>
- Energy (cal) = 6.85×10⁵ cal
- Energy (KJ) =?
<h3>Conversion scale </h3>
1 cal = 0.004184 KJ
<h3>How to convert 6.85×10⁵ cal to kilojoules</h3>
1 cal = 0.004184 KJ
Therefore,
6.85×10⁵ cal = 6.85×10⁵ × 0.004184
6.85×10⁵ cal = 2866.04 KJ
Thus, 6.85×10⁵ cal is equivalent to 2866.04 KJ
Learn more about conversion:
brainly.com/question/2139943
