Answer:
1) The concentration of HCl = 0.1 M.
2) The table that can be used to organize the information correctly is C.
Explanation:
<u><em>1) The concentration of HCl:</em></u>
- We know that the no. of millimoles of the acid is equal to the no. of millimoles of the base at the neutralization point.
which means that: <em>(MV)HCl = (MV)NaOH,</em>
M of HCl = ??? M, V of HCl = 25.0 mL.
M of NaOH = 0.2 M, V of NaOH = 12.5 mL.
∴ M of HCl = (MV)NaOH/V of HCl = (0.2 M)(12.5 mL)/(25.0 mL) = 0.1 M.
<em>2) The table that can be used to organize the information correctly is C</em>
<em></em>
<em>Table A and B are the same and reported volume of HCl and NaOH is wrong.</em>
<em>Table C is right, contain the correct volumes and concentration of NaOH and missed the concentration of HCl which is 0.1 M.</em>
<em>Table D reported the volume and the concentration of HCl wrongly and also the concentration of NaOH. The data reported of HCl and NaOH is reversed.</em>
The enthalpy of fusion of a substance is the energy required to change the state of a substance from solid to liquid at a constant temperature. The enthalpy of fusion for iodine, I₂, will be higher. This is because there are stronger intermolecular forces holding the iodine molecules together. The stronger intermolecular forces arise from the fact that iodine is a much larger molecule, so it has much more electrons resulting in higher Van der Waal's forces. This is also visible in the fact that at room temperature, iodine is a solid while nitrogen is a gas.<span />
Answer: 568g/mol
Explanation:
It should be noted that there are 40 atoms of carbon in lycopene.
Since mass of 1 carbon = 12g/mol
Mass of 40 carbon atoms = 40 × 12g/mol = 480g/mol
Let the molar mass of lycopene be represented by x.
Therefore the molar mass of carbon = x × mass percent of carbon in lycopene
x × 84.49% = 480g/mol
x × 0.8449 = 480g/mol
x = 480/0.8449
x = 568g/mol
The molar mass of lycopene is 568g/mol
A chemical equation should be balanced, so that when you make calculations based on the equation you must be able to relate the products to the reactants or vice versa. An example of using equations for calculations in chemistry is with the subject stoichiometry.
The collagen provides structural support.
