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Answer:48kg of SiO2, 0.5kg of Al2O3, and 1.5kg of B2O3
Will be the final product
Explanation:
I) 96wt% of SiO2 will amount to 96/100*50 = 0.96*50=48kg of SiO2
ii) 1wt% of Al2O3 will amount to 1/100*50 = 0.01*50=0.5kg of Al2O3
III) 3wt% of B2O3 will amount to 3/100*50 = 0.03*50=1.5kg of B2O3..
The overall product form 48+ 0.5+1.5= 50kg
There are 0.566 moles of carbonate in sodium carbonate.
<h3>CALCULATE MOLES:</h3>
- The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.
- no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3
- Molar mass of Na2CO3 = 23(2) + 12 + 16(3)
- = 46 + 12 + 48 = 106g/mol
- mass of CO3 = 12 + 48 = 60g
- no. of moles of CO3 = 60/106
- no. of moles of CO3 = 0.566mol
- Therefore, there are 0.566 moles of carbonate in sodium carbonate.
Learn more about number of moles at: brainly.com/question/1542846
Carbon has 4 valence electrons, oxygen has 6
Mole ratio for the reaction is 1:1
no of moles in NaOH that reacted= 1*21.17/1000=0.02117mols
molarity of HCl=0.02117*10/1000
=2.117M
