What is the mass in grams of 0.250 mol of the common antacid calcium carbonate?

Molar mass CaCO3 = 100 g/mol

Explanation:

So if we have 75 four or 5 g of the sample And there is 25.8 Percent calcium carbonate by mass. That's 25.8 g of calcium carbonate In 100 g of the sample. I'll do the most application at the end. Um And then if we know the grams of calcium carbonate, well, just reverse the conversion factor. Yeah, 100 g of the sample contains 25 0.8 grams of calcium carbonate. And when we do this, multiplication division, for part a 75.45 Times 25.8, divided by a 1000 Uh 23 significant figures is 1.9 five g of calcium carbonate. Mhm. And then for part B 18.8 times 100 Divided by 25.8 is 2. 3 significant figures 72 0.9 grams. Mhm of the sample. Yeah.

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol