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dusya [7]
2 years ago
11

What is the engine piston displacement in liters of an engine who’s displacement is listed as 430 in.²

Chemistry
1 answer:
sergeinik [125]2 years ago
6 0

For an engine whose displacement is listed as 430 in.², the engine piston displacement in liters is mathematically given as

PD= 7.37 L  

<h3>What is the engine piston displacement in liters of an engine whose displacement is listed as 430 in.²?</h3>

Generally, the equation for the dimensional analysis method,\, we convert in to L is mathematically given as

l*(v/l)*l/v

Therefore,  piston displacement

PD=(450 in3) . (1 dm3 / 61.024 in3) . (1 L / 1 dm3)

PD= 7.37 L  

In conclusion, piston displacement

PD= 7.37 L  

Read more about volume

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How would you prepare a 1 L solution of 3 M MgO?
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Answer:  175.35g

Explanation:  A 3 M solution has 3 moles of solute per litre.

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8 0
3 years ago
How many grams of K2O will be produced from 0.50 g of K<br> and 0.10 g of O2?
Rudik [331]

Answer:

0.6g

Explanation:

Given parameters:

Mass of K = 0.5g

Mass of O₂  = 0.10g

Unknown:

Mass of K₂O  = ?

Solution:

To solve this problem, let us write the reaction equation first;

                   4K   +   O₂     →     2K₂O

The reaction above delineates the balanced chemical reaction.

To solve this problem, we need to know the limiting reactant. This reactant is the one that determines the amount and extent of the reaction because it is given in short supply. The other reactant is the one in excess.

Start off by find the number of moles of the reactant;

     Number of moles =  \frac{mass}{molar mass}

         Molar mas of K  = 39g/mol

          Molar mass of O₂   = 2(16) = 32g/mol

 Number of moles of K  = \frac{0.5}{39}   = 0.013moles

 Number of moles of O₂    = \frac{0.1}{32}   = 0.031moles

From the balanced reaction;

          4 moles of K reacted with 1 mole of O₂

         0.013 moles of K will react with \frac{0.013}{4}   = 0.0078 moles of O₂

We see that oxygen gas is in excess. We were given 0.031moles of the gas but only require 0.0078moles of oxygen gas.

The limiting reactant is potassium.

    therefore;

              4 moles of K produced 2 moles of K₂O

             0.013 moles of K will produce \frac{0.013 x 2}{4}   = 0.0065‬moles of K₂O

to find the mass of K₂O;

   Mass of K₂O  = number of moles x molar mass

                Molar mass of K₂O  = 2(39) + 16  = 94g/mol

  Mass of K₂O = 0.0065 x 94  = 0.6g

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3 years ago
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