Answer:
Name Formula and Charge Name Formula and Charge
ammonium NH4+ hydroxide OH−
acetate C2H3O2−, or CH3COO− nitrate NO3−
bicarbonate (hydrogen carbonate) HCO3− nitrite NO2−
bisulfate (hydrogen sulfate) HSO4− peroxide O22−
carbonate CO32− perchlorate ClO4−
chlorate ClO3− phosphate PO43−
chromate CrO42− sulfate SO42−
cyanide CN− sulfite SO32−
dichromate Cr2O72− triiodide I3−
Explanation:
What are the processes we are supposed to be classifying?
The concept of water by displacement. D=m/v therefore m=15.5 g and volume=5 mL (final - initial) then calculate the density
Answer:
Molarity: 0.522M
Percentage by mass: 2.36 (w/w) %
Explanation:
Formic acid, HCOOH reacts with NaOH as follows:
HCOOH + NaOH → NaCOOH + H₂O
To solve this question we must find the moles of NaOH added = Moles formic acid. Taken into account the dilution that was made we can find the moles -And molarity of formic acid and its percentage by mass as follows:
<em>Moles NaOH = Moles HCOOH:</em>
0.01580L * (0.1322mol / L) =0.002089 moles HCOOH
<em>Moles in the original solution:</em>
0.002089 moles HCOOH * (25mL / 10mL) = 0.005222 moles HCOOH
<em>Molarity of the solution:</em>
0.005222 moles HCOOH / 0.01000L =
<h3>0.522M</h3>
<em>Mass HCOOH in 1L -Molar mass: 46.03g/mol-</em>
0.522moles * (46.03g / mol) = 24.04g HCOOH
<em>Mass solution:</em>
1L = 1000mL * (1.02g / mL) = 1020g solution
<em>Mass percent:</em>
24.04g HCOOH / 1020g solution * 100
2.36 (w/w) %
Answer:
Kc = 12.58
Explanation:
Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3
Kc = (0.052441)(0.10513)/(0.002385)(0.18373)
Kc = 0.0005513/0.000438
Kc = 12.58
Hope that helps!!
