Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten.
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.
Answer:
MnO- Manganese Oxide
Explanation:
Empirical formula: This is the formula that shows the ratio of elements
present in a
compound.
How to determine Empirical formula
1. First arrange the symbols of the elements present in the compound
alphabetically to determine the real empirical formula. Although, there
are exceptions to this rule, E.g H2So4
2. Divide the percentage composition by the mass number.
3. Then divide through by the smallest number.
4. The resulting answer is the ratio attached to the elements present in
a compound.
Mn O
% composition 72.1 27.9
Divide by mass number 54.94 16
1.31 1.74
Divide by the smallest number 1.31 1.31
1 1.3
The resulting ratio is 1:1
Hence the Empirical formula is MnO, Manganese oxide
Answer: For transverse waves, the waves move in perpendicular direction to the source of vibration.
For longitudinal waves, the waves move in parallel direction to the source of vibration .
They are similar within the sense that energy is transferred within the kind of waves.
Explanation:
Answer:
B. –99 kJ.
Explanation:
We have the following information:
1. C(s) + O₂(g) → CO₂(g);
ΔH = -393 kJ
2. 2CO(g) + O₂ → 2CO₂(g);
ΔH = -588 kJ
Using Hess's Law, Our target equation has C(s) on the left hand side, so we re-write equation 1:
1. C(s) + O₂(g) → CO₂(g);
ΔH = -393 kJ
So, we reverse equation 2 and divide by 2, we have equation 3:
3. CO₂(g) → CO(g) + ½O₂;
ΔH = +294 kJ
That is, change the sign of ΔH and divide by 2. Then we add equations 1 and 3 and their ΔH values.
This gives:
C(s) +½O₂(g) → CO(g);
ΔH = +294 - 393 kJ
= -99 kJ
The standard enthalpy of formation of carbon monoxide is -99 kJ/mol.
