Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
Answer:
22.4 L at standard temperature and pressure.
Answer:
0.5M is the answer.
Explanation:
1M solution is the solution containing 1mole solute dissolved per litre of solution.
Using unitary method,
1000cc gives 1M.
1cc gives 1/1000M.
500 cc gives 500/1000M=0.5M
I believe the answer is <span>can be elements or compounds
In this case, elements in the decomposition reaction is the substance that cannot be separated into simpler substances.
Compounds, technically act as a reactant in the decomposition reaction, but since the reaction breakdown one substance into two or more, sometimes it exists in the product.</span>
0.012moldm⁻³
Explanation:
Given parameters:
Mass of AgNO₃ = 1000mg
Volume of water = 500mL
Unknown:
Molarity of solution = ?
Solution:
The molarity of a solution is the number of moles of a solute dissolved in volume of solvent.
Molarity = 
Number of moles of AgNO₃ = ?
Number of moles = 
Molar mass of AgNO₃ = 108 + 14 + 3(16) = 170g/mol
convert mass to g;
1000mg = 1g
Number of moles =
= 0.00588moles
convert the given volume to dm³;
1000mL = 1dm³;
500mL = 0.5dm³
Now solve;
Molarity =
= 0.012moldm⁻³
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Molarity brainly.com/question/9324116
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