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shtirl [24]
2 years ago
12

Write the net ionic equation for the reaction that occurs, if any, when gold is placed in a solution of hydrobromic acid

Chemistry
1 answer:
Gre4nikov [31]2 years ago
3 0

Answer:

ionic salt MX2

Explanation:

Atoms from an element in group IIA are allowed to react with atoms from an element in group VIIA. What type of compound is likely to form

elements from IIA are alkali earth metals (M) and need to lose

2 e^1- to become stable as M^2+

elements from VIIA ar halogens (X) and need to gain 1 e^1-

to become stable so one M^2+ combines with 2 X^1- to make

an ionic salt MX2

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the density of aluminum is 2.70g/cm^3. a piece of aluminum foil has a volume of 54.0 cm^3. what is the mass of this piece of alu
Neporo4naja [7]
The  mass  of  aluminium  foil   is   calculated   as  follows
mass  =  density  x  volume
density =  2.70  g/cm^3
volume  54  cm^3
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3 years ago
Which compound is used to make asphalt?
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3 years ago
5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H
cricket20 [7]
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572.  That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

7 0
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