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2 years ago

5

whick atom do you think from the basic structure of olympicene ? think of the elementvthat form the basic of life

1 answer:

Sloan [31]2 years ago

6 0

Answer:

carbon

Explanation:

Olympicene is an organic carbon based molecule formed of five rings, of which four are benzene rings, joined in the shape of the Olympic rings.

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A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

-BARSIC- [3]

Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

$V = \frac{ 4}{3 }\pi r^{3} = \frac{ 4}{3 }\pi\times (\text{1150 km})^{3} = 6.37 \times 10^{9} \text{ km}^{3}$

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

<em>B) Radius</em>

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

3 0

3 years ago

A _ is a whole number that appears before formula in an equation

Marina CMI [18]

A coefficient is a whole number that appears before the formula in an equation.

7 0

3 years ago

For the reaction PCl5(g) + heat PCl3(g) + Cl2(g), what will happen when the temperature is decreased

vovikov84 [41]

There will be a shift towards the reactants

5 0

3 years ago

Read 2 more answers

Which of the following is not equal to 42.7 millimeters?

Stella [2.4K]

The answer is: C. 0.00427 m.

A) 1 km = 1000000 mm.

d = 0.0000427 km · 1000000 mm/km.

d = 47.7 mm.

B) 1 hm = 100000 mm.

d = 0.000427 hm · 100000 mm/hm.

d = 42.7 mm.

C) 1 m = 1000 mm.

d = 0.00427 m · 1000 mm/m.

d = 4.27 mm.

D) 1 cm = 10 mm.

d = 4.27 cm · 10 mm/cm.

d = 42.7 mm.

Millimeter (abbreviated: mm, a thousandth part of metar) is an unit of distance in the metric system.

8 0

3 years ago

About 6 × 109 g of gold is thought to be dissolved in the oceans of the world. If the total volume of the oceans is 1.5 × 1021 L

Lelechka [254]

First, determine the number of moles of gold.

Number of moles = $\frac{given mass in g}{molar mass}$

Given mass of gold = $6 \times 10^{9} g$

Molar mass of gold = 196.97 g/mol

Put the values,

Number of moles of gold = $\frac{6 \times 10^{9} g}{196.97 g/mol}$

= $0.03046\times 10^{9} mole$ or $3.046\times 10^{7} moles$

Now, molarity = $\frac{moles of solute}{volume of the solution in liters}$

Put the values, volume of ocean = $1.5 \times 10^{21} L$

Molarity = $\frac{3.046\times 10^{7} moles}{1.5 \times 10^{21} L}$

= $2.03\times 10^{-14} M$

Thus, average molar concentration = $2.03\times 10^{-14} M$

6 0

2 years ago