Answer:
24 atoms makes altogether a molecule of glucose
0.370 mol metal oxide = 55.45 g
<span>1 mol = 55.45/0.370 = 149.86 g </span>
<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>
<span>there is 48/149.86 * 100% O in the sample </span>
<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
Answer:
a. 5.9 × 10⁻³ M/s
b. 0.012 M/s
Explanation:
Let's consider the following reaction.
2 N₂O(g) → 2 N₂(g) + O₂(g)
a.
Time (t): 12.0 s
Δn(O₂): 1.7 × 10⁻² mol
Volume (V): 0.240 L
We can find the average rate of the reaction over this time interval using the following expression.
r = Δn(O₂) / V × t
r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s
r = 5.9 × 10⁻³ M/s
b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:
5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s
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
Answer:
Reducing sugars are absent
Explanation:
Benedict's solution is an substance used in testing sugars. It is mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It can be used instead of Fehling's solution in testing for the presence of reducing sugars.
Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.
