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ANTONII [103]
3 years ago
10

The equation represents the combustion of sucrose. C12H22O11 + 12O2 12CO2 + 11H2O If there are 10.0 g of sucrose and 8.0 g of ox

ygen, how many moles of sucrose are available for this reaction?
Chemistry
1 answer:
Vaselesa [24]3 years ago
7 0
<span>0.0292 moles of sucrose are available. First, lookup the atomic weights of all involved elements Atomic weight Carbon = 12.0107 Atomic weight Hydrogen = 1.00794 Atomic weight Oxygen = 15.999 Now calculate the molar mass of sucrose 12 * 12.0107 + 22 * 1.00794 + 11 * 15.999 = 342.29208 g/mol Divide the mass of sucrose by its molar mass 10.0 g / 342.29208 g/mol = 0.029214816 mol Finally, round the result to 3 significant figures, giving 0.0292 moles</span>
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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
In which example is matter changing
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What is the oxidation number of Cl in chlorate ion ClO3?
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The correct answer to this question is "5." the oxidation number of cl in ClO3 will be a positive 5 because oxygen is naturally a -2 charge. times that by three and then account for the negative charge of the CLO3- ion.

8 0
3 years ago
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Ethylene glycol (C2H6O2) is used as an antifreeze in cars. If 250 g of ethylene glycol is added to 3.00 kg of water, what is the
Zepler [3.9K]

Answer:

2,909 M

Explanation:

molair mass is of.ethylene is 26,04 g/mol

first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.

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Next you need to calculate the amount of moles:

250 g / 26,04 g/mol = 9,60 mol

Now you can calculate the molarity:

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3 years ago
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