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Vlada [557]
3 years ago
14

Covert 0.2598kl to l

Chemistry
1 answer:
Eva8 [605]3 years ago
7 0
Kl in this problem means kilo liter. 1 kilo liter is equivalent to 1000 liter and equal to one cubic meter. 

But in this problem, you are asking to convert this to liter.

So:

0.2598 kl = 1000 liter

                   --------------

                    1 kilo liter

So multiply, we get 259.8 liters or 260 liters. 
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The spaceship does not have any gravitational pull. A spaceship can not produce its own gravity like Earth does.
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Diagram of mass spectroscopy​
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Explanation:

“The basic principle of mass spectrometry (MS) is to generate ions from either inorganic or organic compounds by any suitable method, to separate these ions by their mass-to-charge ratio (m/z) and to detect them qualitatively and quantitatively by their respective m/z and abundance

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Which characteristic will the animal in the image pass on to its offspring<br> Help me out please
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2 years ago
Which conversion factor do you use first to calculate the number of grams of CO 2 produced by the reaction of 50.6 g of CH 4 wit
Brilliant_brown [7]

Answer:

Thus, first conversion of mass of methane into moles by dividing it with 16.04 g/mol

Mass =  138.63 g

Explanation:

The balanced chemical reaction is shown below:-

CH_4+2O_2\rightarrow CO_2+2H_2O

Firstly the moles of methane gas reacted must be calculate as:-

Given, mass of methane = 50.6 g

Molar mass of methane gas = 16.04 g/mol

The formula for the calculation of moles is:-

Moles=\frac{Mass\ taken}{Molar\ mass}=\frac{50.6}{16.04}\ mol=3.15\ mol

Thus, from the reaction stoichiometry,

1 mole of methane produces 1 mole of carbon dioxide

Also,

3.15 mole of methane produces 3.15 mole of carbon dioxide

Moles of CO_2 = 3.15 mole

Molar mass of CO_2 = 44.01 g/mol

Mass = Moles*Molar mass = 3.15\times 44.01 g = 138.63 g

3 0
3 years ago
Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4 NH3 + 5 O2 → 4 NO + 6 H2O When 40.0 g NH3 and 50.0 g
luda_lava [24]

Answer:

18.75 g of NH3.

Explanation:

The balanced equation for the reaction is given below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the masses of NH3 and O2 that reacted from the balanced equation.

This can be obtained as follow:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160 g

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Next, we shall determine the excess reactant. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, 40 g of NH3 will react with = (40 × 160)/68 = 94.12 g of O2.

From the calculations made above, we can see that it will take a higher amount of O2 i.e 94.12g than what was given i.e 50g to react completely with 40 g of NH3.

Therefore, O2 is the limiting reactant and NH3 is the excess reactant.

Next we shall determine the mass of excess reactant that reacted. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, Xg of NH3 will react with 50 g of O2 i.e

Xg of NH3 = (68 × 50)/160

Xg of NH3 = 21.25 g

Therefore, 21.25 g of NH3 (excess reactant) were consumed in the reaction.

Finally, we shall determine mass of the remaining excess reactant as follow:

Mass of excess reactant = 40 g

Mass of excess reactant that reacted = 21.25 g

Mass of excess reactant remainig =?

Mass of excess reactant remainig = (Mass of excess reactant) – (Mass of excess reactant that reacted)

Mass of excess reactant remainig

= 40 – 21.25

= 18.75 g

Therefore, the mass of excess reactant remaining is 18.75 g of NH3.

8 0
3 years ago
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