Answer:
the law of conservation of matter.
Explanation:
I am in chem
Answer:
c. add coefficients as needed
Explanation:
A chemical equation is defined as the equation that shows changes in a chemical reaction. A chemical equation consist of reactant and product, reactant is at left side of the arrow and product is at right side of the arrow.
Reactant => Product
While balancing a chemical equation, the basic rule is to balance the coefficient as required. Coefficient represents the number of molecules and is used at front of a chemical symbol. Change in coefficient helps balance the number of atoms or molecules of the substances on both the sides of the arrow.
Subscripts are never allowed to change because it can change the chemical involved in the reaction.
Hence, the correct answer is "c. add coefficients as needed".
Moles of H2SO4= 7.5x10^23/ 6.02x10^23 = 1.25 (3sf) moles of H2SO4
Mass of 1 mole of H2SO4= 98.1g
Therefore mass of 7.5x10^23 molecules of H2SO4= 122.63g
Answer:
159 mg caffeine is being extracted in 60 mL dichloromethane
Explanation:
Given that:
mass of caffeine in 100 mL of water = 600 mg
Volume of the water = 100 mL
Partition co-efficient (K) = 4.6
mass of caffeine extracted = ??? (unknown)
The portion of the DCM = 60 mL
Partial co-efficient (K) = 
where; 
solubility of compound in the organic solvent and 
= solubility in aqueous water.
So; we can represent our data as:
÷ 
Since one part of the portion is A and the other part is B
A+B = 60 mL
A+B = 0.60
A= 0.60 - B
4.6=
÷ 
4.6 = 
4.6 × =
4.6 B 
= 0.6 - B
2.76 B = 0.6 - B
2.76 + B = 0.6
3.76 B = 0.6
B = 
B = 0.159 g
B = 159 mg
∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.
Answer:
the compound contains C, H, and some other element of unknownidentity, so we can’t calculate the empirical formula
Explanation:
Mass of CO2 obtained = 3.14 g
Hence number of moles of CO2 = 3.14g/44.0 g = 0.0714 mol
The mass of the carbon in the sample = 0.0714 mol × 12.0g/mol = 0.857 g
Mass of H2O obtained = 1.29 g
Hence number of moles of H2O = 1.29g/18.0 g = 0.0717 mol
The mass of the carbon in the sample = 0.0717 mol × 1g/mol = 0.0717 g
% by mass of carbon = 0.857/1 ×100 = 85.7 %
% by mass of hydrogen = 0.0717/1 × 100 = 7.17%
Mass of carbon and hydrogen = 85.7 + 7.17 = 92.87 %
Hence, there must be an unidentified element that accounts for (100 - 92.87) = 7.13% of the compound.
