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iVinArrow [24]
3 years ago
15

Write the bond line formula for the compound (CH3)2CHCH2C(CH3)3.​

Chemistry
1 answer:
maks197457 [2]3 years ago
3 0

Formula is (CH3)2CHCH2C(CH3)3 and the name is 2,2,4-trimethylpentane

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A and B are experiencing winter. The picture which isn't available here in this question is attached below.

Option C.

<h3><u>Explanation:</u></h3>

The earth is tilt by an angle of 23.2° to the vertical plane. This makes the seasonal variation of earth, because in some time of the year, the northern hemisphere faces the sun directly, experiencing summer and then southern hemisphere is away from summer experiencing winter and vice versa. The summer occurs when the place directly faces the sun. And the winter happens when the place obliquely faces the sun or doesn't face the sun at all.

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The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P
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Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

  • Pb: 207.2;
  • S: 32.06.

To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is \rm PbS. There will be 890 grams of \rm PbS in one kilogram of this rock.

Formula mass of \rm PbS:

M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}.

How many moles of \rm PbS formula units in that 890 grams of \rm PbS?

\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol.

There's one mole of \rm Pb in each mole of \rm PbS. There are thus \rm 3.71980\; mol of \rm Pb in one kilogram of this rock.

What will be the mass of that \rm 3.71980\; mol of \rm Pb?

m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg.

In other words, the \rm PbS in 1 kilogram of this rock contains \rm 0.770743\; kg of lead \rm Pb.

How many kilograms of the rock will contain enough \rm PbS to provide 1.5 kilogram of \rm Pb?

\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg.

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