All categories

3 years ago

one horse power is a unit of power equal to 746W. Jopw much engery can a 150-horsepower engine transform in 10 secs?

Chemistry

1 answer:

frez [133]3 years ago

4 0

Answer:

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

(150 horsepower) x (746 watt/HP) x (1 joule/sec / watt) x (10 sec)

= (150 x 746 x 1 x 10) joule

= 1,119,000 joules .

Explanation:

Pls mark the brainliest

You might be interested in

19. Which of the following best describes the role of the spark from the spark plug in an automobile engine?

sergey [27]

The role of a spark plug is to supply some of the energy of activation for the combustion reaction.

Explanation:

A Spark plug is a tiny bolt of lightning in which a spark of electricity is emitted across a gap creating the ignition of the combustion chamber thereby starting the engine. By putting an engine piston in motion we can power up which produces a smooth burn of the compressed air-fuel mixture.

An electrical device that fits into the cylinder head and ignites compressed aerosol gasoline by an electric spark. They have an insulated electrode connected to a coil that ignites thereby producing sparks.
The spark plug works as a heat exchanger. They tend to pull unwanted thermal energy from the combustion chamber and heat is transferred to the engine's cooling system. Thus they supply some of the energy for the activation of engines.

4 0

4 years ago

Carbon dioxide (CO2) is a gaseous compound. Calculate the percent composition of this compound. Answer using three significant f

Tamiku [17]

Carbon dioxide is a gaseous molecule made up of the elements, C and O. Each mole of carbon dioxide has one mole C and two mole oxygen atoms.

Molar mass of carbon dioxide ( $CO_{2}$ )= $(1 * 12.01\frac{g}{mol}) +(2* 16\frac{g}{mol})=44.01\frac{g}{mol}$

Percentage by mass of carbon = $\frac{(1*12.01\frac{g}{mol}) }{44.01\frac{g}{mol} } *100=27.3%$

Percentage by mass of oxygen = $\frac{(2*16\frac{g}{mol})}{44.01\frac{g}{mol} } *100=72.7%$

Therefore C is 27.3 % and O is 72.7 % by mass in 1 mol CO $_{2}$

7 0

4 years ago

Read 2 more answers

If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×

Elodia [21]

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

3 0

4 years ago

A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water

Taya2010 [7]

Answer:

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

So the sample will contain 2.912 - ( .347 + 2.1 ) g of oxygen .

= .465 g of oxygen .

moles of Carbon = 2.1 / 12 = .175

moles of hydrogen = .347 / 1 = .347

moles of oxygen = .465 / 16 = .029

Ratio of moles of carbon , oxygen and hydrogen ( C,O,H )

= 0.175 : 0.029 : 0.347

= .175/ .029 : 1 : .347 / .029

= 6 : 1 : 12

So empirical formula = C₆H₁₂O

Let the molecular formula be $(C_6H_{12}O)_n$

molecular weight = n ( 6 x 12 + 12x 1 + 16)

= 100 n

Given 100 n = 100.1

n = 1

Molecular formula = C₆H₁₂O.

3 0

3 years ago

The thermite reaction reacts iron (III) oxide, Fe2O3, with aluminium powder,Al, the form aluminium oxide, Al2O3 and iron, Fe.

lara [203]

Answer:

This answer assumes that the strated "16.0g of iron" was meant to be 16.0 grams of iron(III) oxide.

Explanation:

To start, the thermite equation must be balanced.

I find:

1Fe2O3 + 2Al = 1Al2O3 + 2Fe

This tells us we need 2 moles of Al for every 1 mole of Fe2O3.

Now calculate the moles of each reactant:

Moles Fe2O3: 16.0 g/159.7 g/mole = 0.100 moles Fe2O3

Moles Al: 8.1 /26.98 g/mole = 0.300 moles Al

The balanced equation says that in order to react all of the Fe2O3 we'd need twice that amount (in moles) of the Al. (0.100 moles Fe2O3)*(2) = 0.200 moles Al.

Which of the two reactants is the limiting reagent?

We have more than enough moles of Al to react with 0.10 moles of Fe2O3. (We have 0.300 moles Al and all we need is 0.200 moles to react with the 0.10 moles of Fe2O3. Fe2O3 is the limiting reagent.

Calculate the maximum mass of iron of iron that could be formed using these quantities of reactants.

The balanced equation tells us that we will obtain 2 moles of Fe for every 1 mole of Fe2O3 consumed. Since Fe2O3 is the limiting reagent, we will assume that it completely reacts. That means 0.1 moles of Fe2O3 is reacted. Since we expect twice that many moles of Fe, we should obtain 0.200 moles of Fe. At 55.85 g/mole, we should obtain:

(0.200 moles Fe)*(55.85 g Fe/mole Fe) = 11.2 grams Fe

5 0

2 years ago