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pychu [463]
3 years ago
5

Considering particles at the subatomic level, carrying out this experiment would help to identify the metals given that: Ca has

the
Choose...
atomic radius. In chemical reactions, it would be
Choose...
for it to lose its valence electrons to form ions. This means it has comparatively
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ionization energies and would react more
Choose...
with the water.
Chemistry
1 answer:
svet-max [94.6K]3 years ago
4 0

Answer:

we can do it again and again and again and again and again and again

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The bowl has more volume, the bearing has more volume. The mass is bigger for the bearing because it is heavier than the bowl. It is made of metal and the weight of it is greater than the bowl.(the body shape).
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Do you think your meal would allow your cells to sufficiently build your cell
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Calcule la normalidad de una solución
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si por favor

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3 years ago
Formula Silver(II) Oxide ​
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Answer:

AgO

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3 years ago
During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

3 0
3 years ago
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