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3 years ago

Fuel is injected at which state diesel cycle

1 answer:

6 0

The Diesel cycle is a combustion process of a reciprocating internal combustion in the cylinders of the engine. In it, fuel is ignited by heat generated during the compression of air in the combustion chamber, into which fuel is then injected and the exhaust from the combustion is pushed out the exhaust pipe or brought out by a turbo

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The melting point of h2o is 0 degrees celsius. this is the same as its:

professor190 [17]

This is the same as its freezing point

hope this helps

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3 years ago

Hook GC lsshvrrjvdjscavsbsgd

Tema [17]

Answer:

I have no idea what that issss

Explanation:

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3 years ago

The chemical equation shows how ammonia reacts with sulfuric acid to produce ammonium sulfate. 2NH3(aq) + H2SO4(aq) (NH4)2SO4(aq

Gemiola [76]

D. 7,930 g ................................ believe me

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3 years ago

Read 2 more answers

3. Determine the enthalpy of formation for propane. 3C(s, gr) + 4H2(g) ---> C3H2(g) CzH3(g) AH = -2219.9 kJ C(s, gr) AH = -39

Mars2501 [29]

Answer:

b I hope this is right good luck

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3 years ago

An aqueous solution is listed as being 33.8% solute by mass with a density of 1.15 g/mL, the molar mass of the solute is 145.6 g

vodomira [7]

Answer:

A) 2.69 M

B) 0.059

Explanation:

A) We have:

33.8% solute by mass= 33.8 g solute/100 g solution

molarity = mol solute/ 1 L solution

molarity= $\frac{33.8 g solute}{100 g solution}$ x $\frac{1.15 g solution}{1 ml}$ x $\frac{1 mol solute}{145.6 g solute}$ x $\frac{1000 ml}{1 L}$

molarity= 2.69 mol solute/L solution = 2.69 M

B) We know that there are 33.8 g of solute in 100 g of solution.

As the total solution is compounded by solute+solvent (in this case, solvent is water), the mass of water is the difference between the mass of the total solution and the mass of solute:

mass of water= 100 g - 33.8 g = 66.2 g

Now, we calculate the number of mol of both solute and water:

mol solute= 33.8 g solute x $\frac{1 mol solute}{145.6 g}$ = 0.232 mol

mol H20= 66.2 g H₂O x $\frac{1 mol H2O}{18 g}$

Finally, the mol fraction of solute (Xsolute) is calculated as follows:

Xsolute= $\frac{mol solute}{total mol}= \frac{mol solute}{mol solute + mol H2O}=\frac{0.232 mol}{0.232 mol + 3.677 mol}$

Xsolute= 0.059

4 0

3 years ago