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write a balanced net ionic equation for the following reaction: BaCl2(aq) + H2SO4 (aq) -- BaSO4(s) + HCl (aq)

Fudgin [204]

The balanced net equation for

BaCl2 (aq) + H2SO4(aq) → BaSO4(s) + HCl (aq) is

Ba^2+(aq) +SO4^2- → BaSO4 (s)

Explanation

Ionic equation is a chemical equation in which electrolytes in aqueous solution are written as dissociated ions.

ionic equation is written using the below steps

Step 1: write a balanced molecular equation

BaCl2 (aq) +H2SO4 (aq)→ BaSO4(s) +2HCl (aq)

Step 2: Break all soluble electrolytes in to ions

= Ba^2+ (aq) + 2Cl^-(aq) + 2H^+(aq) + SO4^2-(aq)→ BaSO4(s) + 2H^+(aq) +2Cl^- (aq)

step 3: cancel the spectator ions in both side of equation ( ions which do not take place in the reaction)

 = 2Cl^- and 2H^+ ions

Step 4: write the final net equation

 Ba^2+(aq) + SO4^2-(aq)→ BaSO4(s)

3 0

3 years ago

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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$

$60mL=0.06L$

Now we can calculate the concentration:

$M=\frac{0.00423mol}{0.06L}$

$M=0.0705$

Now, we can calculate the pOH (to calculate the pH), so:

$pOH=-Log(0.0705)$

$pOH=1.15$

Now we can calculate the pH value:

$14=~pH~+~pOH$

$pH=14-1.15=12.84$

8 0

3 years ago

The molar mass of Cr(OH)2 is:

son4ous [18]

The molar mass of $Cr(OH)_2$ is 86.02 g/mole .

<h3> Explanation: </h3>

The molar mass of a chemical compound is represented as the mass of a unit of that compound separated by the number of substances in that unit, measured in moles. The molar mass is a volume, not molecular, the property of a substance.

The molar mass is a percentage of various examples of the compound, which usually change in mass due to the appearance of isotopes.

From the below attached table, the Molar mass of $Cr(OH)_2$ is 86.0108 g/mol.