Answer:
1.43 (w/w %)
Explanation:
HCl reacts with NH3 as follows:
HCl + NH3 → NH4+ + Cl-
<em>1 mole of HCl reacts per mole of ammonia.</em>
Mass of NH3 is obtained as follows:
<em>Moles HCl:</em>
0.02999L * (0.1068mol / L) = 3.203x10-3 moles HCl = <em>Moles NH3</em>
<em>Mass NH3 in the aliquot:</em>
3.203x10-3 moles NH3 * (17.031g / mol) = 0.0545g.
Mass of sample + water = 22.225g + 75.815g = 98.04g
Dilution factor: 98.04g / 14.842g = 6.6056
That means mass of NH3 in the sample is:
0.0545g * 6.6056 = 0.36g NH3
Weight percent is:
0.36g NH3 / 25.225g * 100
<h3>1.43 (w/w %)</h3>
Molar mass Pb = 207.2 g/mol
1 mole Pb ------------- 207.2
? mole Pb ------------ 9.51 x 10³
moles = 9.51 x 10³ * 1 / 207.2
moles = 9.51 x 10³ / 207.2
= 45.89 moles
hope this helps!
Answer: 173 g ( 0.17 kg in right accuracy)
Explanation: Amount in moles is n = N/Na = 2.0·10^24 / 6.022·10^23 (1/mol).
n = 3.32116 mol. M(Cr) = 52.00 g/mol and mass m = nM = 172.7 g
This answer is 24 because 2.17 x 10 -8 is 24 so that would be your answer
Answer:
I got the answers but it won't let me post it correctly on here....
Explanation:
9.) 10-2.76 =0.0174 [H30+]= 1.74*10-3 M
10.)10-3.65=0.00224 [H3O+] =2.24*10-2 M
11.)10-3.65=0.00224 [OH-]= 2.224*10-4M
12.)10-6.87=0.00000135 [OH-]= 1.35*10-7M
