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baherus [9]
3 years ago
7

How many grams of Ba(OH)2 will dissolve in 100g of water at 75 degree Celsius

Chemistry
1 answer:
netineya [11]3 years ago
3 0
Solubility is a function of temperature. But for this case solubility is not given at that certain temperature and it is somewhat difficult to find for certain datas. I would be using the solubility at 20 degrees celsius which is 1.85g/100g Water

.0185 g / g water = g barium hydroxide / 100 g

1.85 g of barium hydroxide can be dissolved in 100 g of water. 

Hope this could help. Have a nice day.
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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
What describes a mixture
kirill [66]

Answer:

a mixture is a material made up of two or more different substances which are physically combined.

4 0
3 years ago
I will mark brainliest
Veronika [31]

I believe the answer is B??????????? Hope this helps

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5 0
3 years ago
Read 2 more answers
List the ions in solution in the electrolysis of conc.sodium chloride<br>​
Licemer1 [7]
Chloride ions Cl –(aq) (from the dissolved sodium chloride) are discharged at the positive electrode as chlorine gas, Cl 2(g) sodium ions Na +(aq) (from the dissolved sodium chloride) and hydroxide ions OH –(aq) (from the water) stay behind - they form sodium hydroxide solution, NaOH(aq)
5 0
3 years ago
The solubility of lead(ii) chloride is 0.45 g/100 ml of solution. what is the ksp of pbcl2? 4.9 × 10-2 1.7 × 10-5 8.5 × 10-6 4.2
Artyom0805 [142]
Answer:
1.7 * 10^-5

Explanation:
1- get the number of moles of PbCl2:
number of moles = mass / molar mass
number of moles = 0.45 / 278.1 = 1.618 * 10^-3 moles

2- get the concentration of Pb2+:
molarity = number of moles of solute / volume of solution in liters
molarity = (1.618 * 10^-3) / (0.1) = 0.0162 M

3- getting concentration of Cl-:
<span>PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq) 
</span>We can note that:
For a certain amount of Pb2+ formed, twice this amount of Cl- is formed.
This means that:
for 0.0162 M of Pb2+, 2*0.0168 = 0.0324 M of Cl- is formed

4- getting Ksp:
Ksp = [Pb2+][Cl-]²
Ksp = (0.0162)*(0.0324)²
Ksp = 1.7 * 10^-5

Hope this helps :)
7 0
4 years ago
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