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3 years ago

15 POINTS FOR ANYONE WHO CAN ANSWER THIS!!!!!! plZ

Chemistry

1 answer:

yaroslaw [1]3 years ago

8 0

Answer:

A. 2Al(s) + 3Cl2(g) —> 2AlCl3(s)

B. 2Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

C. H2(g) + Br2(l) —> 2HBr(g)

Explanation:

A. Reaction between aluminium metal and chlorine gas.

Al(s) + Cl2(g) —> AlCl3(s)

The above equation can be balance as follow:

There are 2 atoms of Cl on the left side and 3 atoms on the right side. It can be balance by putting 3 in front of Cl2 and 2 in front of AlCl3 as shown below:

Al(s) + 3Cl2(g) —> 2AlCl3(s)

There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Al as shown below:

2Al(s) + 3Cl2(g) —> 2AlCl3(s)

Now the equation is balanced.

B. Reaction between lithium metal and liquid water.

Li(s) + H2O(l) —> LiOH(aq) + H2(g)

The above equation can be balance as follow:

There are 2 atoms of H on the left side and a total of 3 atoms on the right side. It can be balance by putting 2 in front of H2O and 2 in front of LiOH as shown below:

Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

There are 2 atoms of Li on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Li as shown below:

2Li(s) + 2H2O(l) —> 2LiOH(aq) + H2(g)

Now, the equation is balanced.

C. Reaction between gaseous hydrogen and liquid bromine.

H2(g) + Br2(l) —> HBr(g)

The above equation can be balance as follow:

There are 2 atoms of H on the left side and 1 atom on the right. It can be balance by putting 2 in front of HBr as shown below:

H2(g) + Br2(l) —> 2HBr(g)

Now the equation is balanced.

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3 years ago

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To allow all the elements or compounds to separate complete.

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3 years ago

Describe porosity and how that affects groundwater infiltration.

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3 years ago

How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94

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Answer : The mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = $380\times 10^{-3}$ $(1ml=10^{-3}L)$

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

$pH+pOH=14\\pOH=14-8.94=5.06$

$pOH=-log[OH^-]$

$5.06=-log[OH^-]$

$[OH^-]=0.00000871=8.71\times 10^{-6}mole/L$

Now we have to calculate the moles of $OH^-$ .

Formula used : $Moles=Concentration\times Volume$

$\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles$