Explanation
Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1
Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1
Statement 1 is correct but Statement 2 is not correct
Statement 1 is not correct but Statement 2 is correct
Lead is a chemical element with atomic number 82 and symbol Pb comes from the Latin word $$plumbum$$. It is a soft and malleable metal with a density exceeding that of most common materials. Mercury is a chemical element with symbol Hg and atomic number 80. It is commonly known as quicksilver and was formerly named $$hydrargyrum$$.
An atomic mass unit (symbolized amu) is defined as precisely 1/12 the mass of an atom of carbon-12. The carbon-12 ($$C-12$$) atom has six protons and six neutrons in its nucleus. In imprecise terms, one amu is the average of the proton rest mass and the neutron rest mass.
Hence, the correct option is $$\text{A}$$
The molar mass of hydrogen ($$H$$) is $$1$$
Molar mass of sulphur ($$S$$) is $$32$$
Molar mass of oxygen ($$O$$) is $$16$$
We can calculate the molar mass of $$H_2SO_4$$ by adding the molar masses of all the atoms in the molecule.
So,
$$molar\ mass\ of\ H_2SO_4 = (2 \times molar\ mass\ of\ H) + (molar\ mass\ of\ S) + (4 \times molar\ mass\ of\ O)$$
= $$(2 \times 1) + (32) + (4 \times 16)$$
= $$98$$
We know that molar mass of a substance is defined as the mass of $$1$$ mol of that substance. So, we can write it as follows:
Mass of $$1$$ mol of $$H_2SO_4$$ = Molar mass of $$H_2SO_4$$ = $$98$$
Hence, mass of $$5$$ moles of $$H_2SO_4$$ = $$5 \times 98$$ = $$490$$
Explanation:
The molecular mass of oxygen ($$O_2$$) is $$32\ grams$$.
$$\therefore$$ $$6.023 \times 10^{23}$$ molecules of oxygen ($$O_2$$) have mass $$=32\ grams$$.
$$\therefore$$ One molecule of oxygen ($$O_2$$) will have mass $$=\dfrac{32}{6.023\times 10^{23}}=5.31\times 10^{-23}\ grams$$.
Final Answer: One molecule of oxygen ($$O_2$$) have mass $$=5.31\times 10^{-23}\ grams$$. Hence option $$A$$ is correct.
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