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Temka [501]
3 years ago
9

A mass of the crucible (g) 88.000g

Chemistry
1 answer:
nlexa [21]3 years ago
4 0
A)5.000g
b)mole=mass/molarmass
             molar mass of H2O=18g
             mass of water=5g
b)mole of water=5.000/18.000=0.2778g
c)mass of salt=2.342g
d)mole of salt=mass/molar mass
                     pure salt is NaCl
                     molar mass of NaCl=58.50g
 d)mole of salt=2.342/58.5=0.040mol
e)mole ratio=0.27:0.04

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The volume of gas occupied 240.0ml when the pressure is 1.20 atm what volume at constant pressure.will the gas occupies when the
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334.89 ml

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7 0
2 years ago
2H2 +O2 = 2H2O balance equation with 5 oxygen
Ivenika [448]

10H₂    +   5O₂ →    10H₂O

Explanation:

This problem deals with balancing of chemical equations. In balancing chemical equations, the law of conservation of mass must be followed. This states that:

   "In a chemical reaction, matter is neither created nor destroyed but transformed from one form to another".

This meaning of this is that; the number of atoms on each side of the expression must be the same.

                    2H₂    +   O₂ →    2H₂O

let us check is the equation above is balanced;

   

                                      2H₂    +   O₂ →    2H₂O

Elements             reactant                product

H                                  4                           4

O                                   2                           2

We can see vividly that the equation is balanced;

Now; if we have 5 oxygen gas, we multiply the equation through by 5:

 

                                 5 x          (  2H₂    +   O₂ →    2H₂O   )

     ⇒        10H₂    +   5O₂ →    10H₂O

Elements             reactant                product

H                                  20                       20

O                                  10                        10

learn more:

Balanced equation brainly.com/question/11102790

#learnwithBrainly

5 0
3 years ago
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
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