3.8mL of 0.42 phosphoric acid is required.
Reaction
2H3PO4 + 3CaCL3 → Ca3(PO)4 + 6HCl
moles CaCl2 =0.16 mol/L x0.010 L = 0.0016 mol
moles of H3Po4
= 0.0016mol of CaCl2 x 2 mole of H3PO4/3mole of CaCl2
= 0.00106 mol
V of H3PO4 = 0.0016/0.42 = 0.0038L = 3.8mL
V of H3PO4=3.8mL
To know more about calculation in milliliters refer to:-
brainly.com/question/23276655
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Multiply the mass by the abundance and add each. Make sure to convert the percentage into a decimal. (49.946 * .043) + (51.941 * .838) + (52.941 * .095) + (53.939 * .024) = 51.99 round up using sig figs and the answer is c. 52.00 amu
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