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elena-14-01-66 [18.8K]
3 years ago
12

The Average Speed of the orbiting space shuttle is

Chemistry
1 answer:
maxonik [38]3 years ago
8 0

Explanation:

It is given that, the Average Speed of the orbiting space shuttle is  17500 miles/hour.

We need to convert the speed in kilometers/ second

We know that,

1 mile = 1.609 km

or

1 km = 0.621 miles

1 hour = 3600 seconds

17500\ \dfrac{\text{miles}}{\text{hour}}=17500\ \dfrac{\text{miles}}{\text{h}}\times \dfrac{1\ h}{3600\ s}\\\\=17500\times \dfrac{\text{miles}}{3600\ s}

Now cancel the miles in numerator.

17500\times \dfrac{\text{miles}}{3600\ s}=17500\times \dfrac{\text{miles}}{3600\ s}\times \dfrac{1.609\ km}{1\ \text{miles}}\\\\=17500\times \dfrac{1.609}{3600}\ km/s\\\\=7.82\ km/s

So, 17500 miles/hour is equal to 7.82 km/s.

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In two or more complete sentences, explain how to balance the combustion equation: ___C5H12 + ___O2 ⟶ ___CO2 + ___H2O
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Why should you prefer dry ashing to wet ashing of food sample such as cabbage​
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9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r
Mumz [18]

Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\  \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\  \\ Percent\text{ }yield=0.9484\times100\% \\  \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}

Therefore, the percent yield for the reaction is 95%.

3 0
1 year ago
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