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1 year ago

A jogger runs at an average speed of 5.9 mi/h.

Chemistry

1 answer:

madreJ [45]1 year ago

4 0

She will cover distance in 3 days and 16 hours .the time will be 3:15 am on the fourth day.

Convert 5.9 miles/hour to feet/hour there are 5,280 feet in a mile

total distance covered = 47500

distance = rate x time

now, time = 47500/5280

time = 88 hours

88 hours = 3 Days and 16 Hours.

now the time will be 3 :15 am on the fourth day.

The overall distance the object covers in a given amount of time is its average speed. A scalar value represents the average speed. It has no direction and is indicated by the magnitude.

The distance traveled divided by the time taken is the most popular formula for calculating average speed. The alternative formula is to add the initial and final speeds together, then divide by 2.

Learn more about distance time problems here: -

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PLEASE HELP! 25 POINTS!!!! I got the #1, just not #2 and #3. An industrial chemical company has opened a new plant that will pro

podryga [215]

Answer :

Part 1 : Balanced reaction, $3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Part 2 : The theoretical yield of $NH_3$ gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of $N_2$ = 475 g

Molar mass of $N_2$ = 28 g/mole

Molar mass of $NH_3$ = 17 g/mole

Experimental yield of $NH_3$ = 397 g

Answer for Part (1) :

The balanced chemical reaction is,

$3H_2(g)+N_2(g)\rightarrow 2NH_3(g)$

Answer for Part (2) :

First we have to calculate the moles of $N_2$ .

$\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles$

From the given reaction, we conclude that

1 moles of $N_2$ gas react to give 2 moles of $NH_3$ gas

16.96 moles of $N_2$ gas react to give $\frac{2}{1}\times 16.96=33.92$ moles of $NH_3$ gas

Now we have to calculate the mass of $NH_3$ gas.

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

$\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g$

Therefore, the theoretical yield of $NH_3$ gas = 440.96 g

Answer for Part (3) :

Formula used for percent yield :

$\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100$

$\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%$

Therefore, the % yield of ammonia is 90.03 %

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3 years ago

What type of bond forms when electrons are transferred (lose or gain) from one atom to another?

Zarrin [17]

........ Ionic bond......

7 0

3 years ago

A LA TEMPERATURA DE 35°C, UNA MUESTRA DE DIOXIDO DE CARBONO OCUPA UN VOLUMEN DE 350 ML. ¿Qué CAMBIO DE VOLUMEN SE PRODUCIRA SI L

Kisachek [45]

Answer:

New volume = 150 mL

Explanation:

Initial temperature, T₁ = 35°C

Initial volume, V₁ = 350 mL

We need to find the change in volume when the temperature drops to 15°C.

The relation between the temperature and the volume is given by Charle's law. Let new volume is V₂. It can be given by :

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So, the new volume is 150 mL.

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3 years ago

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Mekhanik [1.2K]

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3 years ago

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Nataly [62]

Answer:

2:8

Explanation:

The reaction equation is a given as:

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From the reaction equation, the mole ratio is 2:8

Butane is C₄H₁₀

Carbon dioxide CO₂

From the reaction;

2 moles of butane will produce 8 moles of carbon dioxide

3 0

3 years ago

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