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Colt1911 [192]
3 years ago
11

An ambulance took 2 hours to cover a distance of 10 kilometers. Find its speed.​

Chemistry
2 answers:
bekas [8.4K]3 years ago
7 0

Answer:

5km/hour

Explanation:

2hours = 10km

/2              /2

1 hour = 5km

balandron [24]3 years ago
3 0
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>:-</h2>

An ambulance took 2 hours to cover a distance of 10 kilometers. Find its speed.

<h2><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>:-</h2>

<h3>Given:-</h3>

Distance = 10 kilometers.

Time = 2 hours.

<h3>To Find:-</h3>

Its speed.

<h2>Solution:-</h2>

We know,

{ \boxed{\bf{\rm \red {Speed \: = \: Distance \: × \: Time}}}}

Speed = 10 kilometers × 2 hours

Speed = 20 km/hr

<h3>Its speed is <u>2</u><u>0</u><u> </u><u>k</u><u>m</u><u>/</u><u>h</u><u>r</u>. [Answer]</h3>
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Select all of the following statements that are false about ΔGo and ΔG:a) If the reaction has a large negative ΔGo value, the re
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Answer:

a) If the reaction has a large negative ΔGo value, the reaction must reach equilibrium at a small extent of reaction value

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Thus; from above mentioned, the statements that are true about ΔG⁰ and ΔG are:

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For a reaction that reaches equilibrium, the minimum value of free energy must be at the equilibrium point

If ΔG⁰ , measured at an extent of reaction = 0.5, is positive, the sign for ΔG when the extent of reaction = 0.80 is also positive.

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Ammonia gas will react with oxygen gas to yield nitrogen monoxide gas and water vapor.
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<u>b) </u> <u>9.50 grams O2</u>

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Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

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Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

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4NH3 + 5O2 → 4NO + 6H2O

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Calculate moles of oxygen = mass O2/ molar mass O2

moles oxygen =  6.73 grams / 32.00 g/mol = 0.210 moles

Calculate moles of NH3

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b. If 6.42g of water is produced, how many grams of oxygen gas reacted?

Calculate moles of H2O = 6.42 grams / 18.02 g/mol = 0.356 moles

Calculate moles of O2:

For 4 moles of NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O

For 0.356 moles H2O we'll need 5/6 * 0.356 = 0.297 moles O2

Calculate mass of O2 = moles O2 * molar mass O2

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c. If the reaction uses up 9.43105 g of ammonia, how many kilograms of nitrogen monoxide will be formed?

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For 0.5538 moles of NH3 we'll have 0.5538 moles NO

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<u />

<u />

d. When 2.51 g of ammonia react with 3.76 g of oxygen, 2.27 g of water vapor are produced. What is the percentage yield of water?

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<em>Calculate moles of O2 </em>= 3.76 grams / 32 g/mol = 0.118 moles

<em>Determine the limiting reactant</em>

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<em>Calculate mass H2O</em> = 0.1416 moles * 18.02 g/mol = 2.552 grams H2O

<em>Calculate % yield</em> = (2.27/2.552)*100 % = <u>88.9 %</u>

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