Answer:
Explanation:
6.90 mol x grams
2As + 6NaOH = 2Na3AsO3 + 3H2
6 mol 2 mol
192 g/mol
6.90 mol NaOH x 2 mol Na3AsO3/6 mol NaOH== 2.3 mol Na3AsO3
2.3 mol Na3AsO3 x 192 g/mol = 442 g Na3AsO3
Answer:
Leroy Jeakins
Explanation:
Catus jack f me in the asss
First, we need to get POH value & we can get it from this formula:
PH + POH = 14
when we have PH = 8.59
∴ POH = 14 - 8.59 = 5.41
then, we need to get [OH-] & we can get it using POH:
POH = -㏒[OH]
5.41 = -㏒[OH]
∴ [OH] = 3.89 x 10^-6
when Ksp = [Fe2-][OH-]^2
and when we have Ksp = 4.87 x 10^-17
so, by substitution:
4.87 x 10^-17 = [Fe2-] * (3.89 x 10 ^-6)^2
∴[Fe2-] = 3.22 x 10^-6 M
∴ above 3.22 x 10^-6 M of Fe2+ , Fe(OH)2 will precipitate
Answer:
1.54 ml
2. 54 g
3. -43.7 °C
4. 97 g
5. 17.74 kJ
Explanation:
1. Volume of water that only came from the melted ice is;
Volume of water plus melted ice-volume of warm water
151 ml - 97 ml = 54 ml
2.Mass of water that came from the melted ice.
Mass=density of water * volume of water from melted ice
M=1 g/ml * 54 ml = 54 g
3. The change in temperature of warm water will be;
Lowest temperature of ice-water mixture-Temperature of warm water
ΔT=0.9-44.6 = -43.7
4. Mass of warm water will be;
Mass of warm water= density of water*volume of warm water
M=1g/ml * 97 = 97 g =0.097 kg
5.
Energy released by warm water as it cools
Q=mcΔT
Given , mass of warm water=m=97
c=4.184 j/g*°C = 4184 J/(kg.C)
Δ= - 43.7
Q=0.097*4184*43.7 =17735.56 J
17735.56/1000 =17.74 kJ