Lead has a density of 11.3 g/cm3 and a mass of 282.5 g. what is its volume?
a.
2.5 cm3
c.
250 cm3
b.
25 cm3
d.
2500 cm3

How many liters of hydrogen gas will be produced at stp from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phosphoric acid (h3po4) the equation is 3mg + 2h3(> mg(po4)2+3h2

To calculate percent by mass, use the equation below: calculate the percent by mass of each element. %n = % %h = % %o = %

Maleic acid is a carbon-hydrogen-oxygen compound used in dyeing and finishing fabrics and as a preservative of oils and fats.* in a combustion analysis, a 1.054-g sample of maleic acid yields 1.599 g of co_2 and 0.327 g of h2o.* when 0.609 g of maleic acid is dissolved in 25.00 g of glacial acetic acid, the freezing point drops by 0.82 c. maleic acid does not ionize in glacial acetic acid, which has a freezing-point depression constant of k_{\rm f}=3.90~^\circ {\rm c}/m.* in a titration experiment, a 0.4250- g sample of maleic acid is dissolved in water and requires 34.03 ml of 0.2152 m koh for its complete neutralization.* the ph of 0.215 g of maleic acid in 50.00 ml of aqueous solution is found to be 1.80.part adetermine the empirical formula for maleic acid. enter the atoms in order ofc, h, and o.part bdetermine the molecular formula for maleic acid. enter the atoms in order of c,h, and o.express your answer as a chemical formula.part cbased on the titration data, how many ionizable h atoms are in each molecule of maleic acid, c4h4o4? express your answer as an integer.i got the part a and b which is cho and c4h4o4 respectively, can anyone me for part c? the below is what i have so far from other qa'sfinding the empirical formula for maleic acid, needs this info"a combustion analysis a 1.054-g sample of maleic acid yields 1.599g of and 0.327g of ".the rest of your data is used to find molar mass, molecular formula, equivalent mass, # of h+ ions / molecule & katypically this should be: a 1.054-g sample of maleic acid yields 1.599g of co2 and 0.327g of h2osofind mass using molar masses1.599g of co2 @ (12.01 g carbon) / (44.01 g co2) = 0.436 grams of carbon0.327g of h2o @ (2.016 g hydrogen) / (18.01 g/mol h2o) = 0.037 grams of hydrogen1.054-g sample - 0.436 grams of carbon - 0.037 grams of hydrogen = 0.581 grams of oxygennow use molar masses to find moles0.436 grams of carbon @ 12.01 g/mol = 0.0363 mol c0.037 grams of hydrogen @ 1.008 g/mol = 0.0367 mol h0.581 grams of oxygen @ 16.00 g/mol = 0.0363 mol oyour empirical formula ischo and molecular is c4h4o4