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3 years ago

10

A runner cruises along at 9 Miles per hour how many days will it take him to run around the World assuming no stop? (40,076km)

1 answer:

Ostrovityanka [42]3 years ago

4 0

Answer:

9x4500=40,500. which is a little more then what you need but it will still be a little more but it will take around 40500 days to run around the world.

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A calorie is another term for carbohydrate.TF

Hunter-Best [27]

<span>This is false. A carbohydrate is a carbon-based molecule that can be utilized by living organisms in order to produce energy. A calorie is a unit of energy often used to measure the amount of energy within food. Another example of energy unit is the Joule, more commonly used within physics.</span>

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3 years ago

Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you

Alja [10]

<u>Answer:</u> The molality of the solution is 0.11 m

<u>Explanation:</u>

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

$0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g$

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (methanol) = 0.135 g

$M_{solute}$ = Molar mass of solute (methanol) = 32 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

$\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m$

Hence, the molality of the solution is 0.11 m

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3 years ago

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klemol [59]

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inysia [295]

The percentage composition of this compound : 40%Ca, 12%C and 48%O

<h3>Further explanation</h3>

Given

20.0 g of calcium,

6.0 g of carbon,

and 24.0 g of oxygen.

Required

The percentage composition

Solution

Total mass of compound :

=mass calcium + mass carbon + mass oxygen

=20 g + 6 g + 24 g

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C-carbon

$\tt \dfrac{6}{50}\times 100\%=12\%$

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2 years ago

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tino4ka555 [31]

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3 years ago