Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles = 
Number of moles = 
= 0.003mole
Answer:
A) oxidizing agent is SO2
B) NaClO is the oxidizing agent
Explanation:
A) This is a redox reaction in which oxidation and reduction occur simultaneously.
Thus, in 2H2S(g) + SO2(g) -> 2H2O(l) + 3S(s);
H2S is reduced as follows;
H2S → S + 2H+ + 2e−
We can see that SO2 has been reduced while H2S gets oxidized since it has changed state from - 2 to 0 . Thus sulphur dioxide is the oxidizing agent.
B) SO2(g) + H2O(l) + NaClO(aq) -> NaCl(aq) + H2SO4(aq)
In this, SO2 undergoes oxidation and NaClO is the oxidizing agent
The answer is Hydrogen
This is what happens when acids react with metals
Answer:
Tyrosine is a polar and aromatic compound. its side chain acidity and basicity is neutral
if a peptide contain only a string of tyrosine residue especially l tyrosine the solubility increases more
Explanation:
even tyrosine number remains constant, tyrosine containing peptide will be more soluble. This peptide is soluble in 1 M HCl (100 mg/ml), with heating. The solubility in water (25 °C) is 0.45 mg/ml in the pH range 3.2 - 7.5.
2.0 mg/ml; at pH 9.5, the solubility is 1.4 mg/ml; and at pH 10, the solubility is 3.8 mg/ml.
Answer:Ian is looking at cells using a microscope. He sees a nucleus and a large vacuole in the central area of a cell. What type of cell is he most likely looking at?
Explanation:
Ian is looking at cells using a microscope. He sees a nucleus and a large vacuole in the central area of a cell. What type of cell is he most likely looking at?
