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3 years ago

12. Chubchub runs a 440m lap in 1.2 minutes. How many laps will Chubchub

Chemistry

1 answer:

Svet_ta [14]3 years ago

3 0

Answer:

the number of laps in the case when he run for 50 minutes is 18,333.33

Explanation:

The computation of the number of laps in the case when he run for 50 minutes is shown below:

Given that

He runs 440m lap in 1.2 minutes

So in 50 minutes he can have laps of

= 440 × 50 ÷ 1.2

= 18,333.33 laps

hence, the number of laps in the case when he run for 50 minutes is 18,333.33

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during the formation of an ionic bond, electrons are transferred from one atom to another true or false

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3 years ago

One of the reactions in a blast furnace used to reduce iron is shown above. How many grams of Fe2O3 are required to produce 15.5

Salsk061 [2.6K]

Answer : The correct option is, (b) 22.1 g

Solution : Given,

Mass of iron = 15.5 g

Molar mass of iron = 56 g/mole

Molar mass of $Fe_2O_3$ = 160 g/mole

First we have to calculate the moles of iron.

$\text{Moles of Fe}=\frac{\text{Mass of Fe}}{\text{Molar mass of Fe}}=\frac{15.5g}{56g/mole}=0.276moles$

Now we have to calculate the moles of $Fe_2O_3$ .

The balanced reaction is,

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

From the balanced reaction, we conclude that

As, 2 moles of iron obtained from 1 mole of $Fe_2O_3$

So, 0.276 moles of iron obtained from $\frac{0.276}{2}=0.138$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{Mass of }Fe_2O_3=\text{Moles of }Fe_2O_3\times \text{Molar mass of }Fe_2O_3$

$\text{Mass of }Fe_2O_3=(0.138mole)\times (160g/mole)=22.08g=22.1g$

Therefore, the amount of $Fe_2O_3$ required are, 22.1 grams.

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3 years ago

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3 years ago

Given the following values for the heats of formation, what is the number of moles of ethane (C2H6, MW 30.0) required to produce

baherus [9]

Answer:

0.641 moles of ethane

Explanation:

Based on the equation:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)

We can determine ΔH of reaction using Hess's law. For this equation:

<em>Hess's law: ΔH products - ΔH reactants</em>

ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}

<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>

<em />

ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}

ΔH = -1559.7kJ/mol

That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:

1.00x10³kJ * (1mole ethane / 1559.7kJ) =

<h3>0.641 moles of ethane</h3>

7 0

3 years ago