One Who Will Answer These Class 9 Chemistry Questions will be marked as a BRAINLIEST:
1. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? The molar mass of Hg and S are 200.6 mol-1 and 32 g mol-1 respectively.
2. (a) Which among the following has the number of molecules:
1 g of hydrogen (H2) or 1 g of methane (CH4)?
(Atomic mass of H = 1 u, C = 12 u)
(b) Calculate the number of particles in 46 g of Na atoms.
(Atomic mass of Na = 23 u)
3. The percentage of three elements, calcium, carbon and oxygen is a sample of calcium carbonate
is given as:
Calcium = 40% ; Carbon = 12%; Oxygen = 48%
If the law of constant proportion is true, what weights of these elements will be present in 1.5 g of another sample of calcium carbonate? [Atomic mass of Ca= 40 u, C= 12u, O = 16 u]
4. During an experiment, the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution.
(a) Are the two solutions of the same concentration?
(b) Compare the mass % of the two solutions.
5. Classify the following as physical or chemical properties:
(a) The composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements.
(b) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas. Zinc reacts with HCl to form ZnCl2 which is soluble in water.
(c) Metallic sodium is soft enough to be cut with a knife.
(d) Most of the metal oxides form bases on interacting with water.
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Answer:
given below
Explanation:
1.)So here mass of HgS given=225 gram.
Molecular mass of HgS is 200.6+32=232.6 gram.
Hence moles present in given mass=225/232.6=0.94 grams.
As you know in 1 mole of HgS there is one mole of Hg.
So in 0.94 moles of HgS there will be 0.94 moles of Hg.
As mass is asked we know mass=moles× molecular mass=0.94×200.6=194.04 gram.
So it exists 194.94 gram Hg in 225 gram HgS.
2(a). 1g of H2 has 0.5×6.022×10^23 molecules.
and 1g of CH4 has 0.0625 ×10^23 molecules.
2(b). the total number of particles are 2×6.022×10^23.
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