Answer:
.3422 M
Explanation:
Mole weight of NaCl ( from periodic table) = 58.439 gm/mole
20 g is how many moles ? 20 g / 58.439 gm/mole=.3422 mole
.3422 mole / 1 liter .3422 M
Result the solution is a basic because .. in there added a strong base.. naoh
Answer:
The answer to your question is 0.22
Explanation:
Data
Acetonitrile (CH₃CN) density = 0.786 g/ml
Methanol (CH₃OH) density = 0.791 g/ml
Volume of CH₃OH = 22 ml
Volume of CH₃CN = 98.4 ml
Process
1.- Calculate the mass of Acetonitrile and the mass of Methanol
density = mass/ volume
mass = density x volume
Acetonitrile
mass = 0.786 x 98.4
= 77.34 g
Methanol
mass = 0.791 x 22
= 17.40 g
2.- Calculate the moles of the reactants
Acetonitrile molar mass = (12 x 2) + (14 x 1) + (3 x 1)
= 24 + 14 + 3
= 41 g
Methanol molar mass = (12 x 1) + (4 x 1) + (16 x 1)
= 12 + 4 + 16
= 32 g
Moles of Acetonitrile
41 g ----------------- 1 mol
77.34g ------------ x
x = (77.34 x 1) / 41
x = 1.89 moles
Moles of Methanol
32 g -------------- 1 mol
17.40 g --------- x
x = (17.40 x 1)/32
x = 0.54 moles
3.- Calculate the mole fraction of Methanol
Total number of moles = 1.89 + 0.54
= 2.43
Mole fraction = moles of Methanol / total number of moles
Mole fraction = 0.54/ 2.43
Mole fraction = 0.22
Answer:
14.91 K.
Explanation:
- To solve this problem, we can use the following relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat transferred to water.
m is the mass of the amount of water (m = 2.0 kg = 2000.0 g).
c is the specific heat capacity of water (c = 4.2 J/g.K).
ΔT is the change in temperature due to the transfer of butane burning.
- To determine Q that to be used in calculation:
Q from 4.000 g of butane is completely burned is - 198.3 kJ = 198300 J.
<em>The negative sign</em><em> symbolizes the the enthalpy change is </em><em>exothermic</em><em>, which means that </em><em>the</em><em> </em><em>energy is released</em><em>.
</em>
- Note that only 63.15% of the energy generated is actually transferred to the water.
∴ Q (the amount of heat transferred to water) = (198300 J)(0.6315) = 125226.45 J.
- Now, we can obtain the change in temperature:
∴ ΔT = Q/m.c. = (125226.45 J) / (2000.0 g)(4.2 J/g.K) = 14.9079 K ≅ 14.91 K.
<em>This means that the temperature is increased by 14.91 K.</em>
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