Speaking of Gen. Chem...

[gwest]

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Just wondering if any wizards could explain any of these questions?? Any help would be appreciated 🙂

what is the mass, in grams of 245 mL of SO2 @ STP?.........................0.694

how much 4M Ca(OH)2 is needed 2 neutralize 300 mL of 3M HNO3?.................112.5 mL

how many atoms of Carbon are in 4.00 E-8 g of C3H8?.....................4E4

if all the chloride in a 5.0g sample of an unknown metal chloride is precipitated as AgCl with 70.9 mL of .201M AgNO3, what is the percent of chloride in the sample?

 

[Notoriousjae]

I'm not a wizard but i'll help work these out. here are the solutions to the first two. for the first one review ideal gas laws and for the second one its a neutralization formula equation.i'll let someone else take a stab at the remaining 2 questions 😀

1. 0.245L * 1mol SO2/22.42L SO2=0.011 mol
0.011mol * 64gSO2/1mol= 0.694g of SO2

2. Neutralization is Volume of acid * Normality of acid =volume of base * Normality of base
So 300mL * 3N = XmL * 8N
XmL= 900/8 = 112.5 mL

 

[armorshell]

Just wondering if any wizards could explain any of these questions?? Any help would be appreciated 🙂

what is the mass, in grams of 245 mL of SO2 @ STP?.........................0.694

how much 4M Ca(OH)2 is needed 2 neutralize 300 mL of 3M HNO3?.................112.5 mL

how many atoms of Carbon are in 4.00 E-8 g of C3H8?.....................4E4

if all the chloride in a 5.0g sample of an unknown metal chloride is precipitated as AgCl with 70.9 mL of .201M AgNO3, what is the percent of chloride in the sample?

3.
4.0 *10^-8g / 44g/mol C3H8 * 6.23*10^23 molecules C3H8/mol * 3 atom C/molecule C3H8 = 1.7*10^15 atoms C.. not sure where 4*10^4 is coming from...

4.
70.0ml AgNO3* 1L/1000ml * .201 mol Ag/L = mols of Chloride = .014 mol

.014 mol Cl * 35.5 g/mol Cl / 5g *100% = 10% Cl

 

[gwest]

thanks for the help!

 

[asckwan]

3.
4.0 *10^-8g / 44g/mol C3H8 * 6.23*10^23 molecules C3H8/mol * 3 atom C/molecule C3H8 = 1.7*10^15 atoms C.. not sure where 4*10^4 is coming from...

4.
70.0ml AgNO3* 1L/1000ml * .201 mol Ag/L = mols of Chloride = .014 mol

.014 mol Cl * 35.5 g/mol Cl / 5g *100% = 10% Cl

Did you do these calculations without a calculator? If so, do you have a quicker way to calculate? It took me a while to divide and multiply out everything.