Which tools do meteorologists use to help create weather forecasts? Choose ONE answer:
2 answers:
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<span>Molar mass (MM) of benzene C6H6
C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x
78x = 468
X = 468/78
X = 6 moles
Molar mass (MM) of Hydrochloric Acid HCl H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol
Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y
36y = 72
Y = 72/36
Y = 2 moles
Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts, Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4
Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1
ANSWER:
</span>
C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x
78x = 468
X = 468/78
X = 6 moles
Molar mass (MM) of Hydrochloric Acid HCl H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol
Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y
36y = 72
Y = 72/36
Y = 2 moles
Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts, Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4
Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1
ANSWER:
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Answer:
(a) 0.699 kJ/K
(b) -0.671 kJ/K
(c) 0.028 kJ/K
Explanation:
The Refrigerant-134a flows into the evaporator as a saturated liquid-vapor mixture and flows out as a saturated vapor at a saturation pressure of 160 kPa and temperature of -15.64°C (estimated from the Saturated Refrigerant-134a Temperature Table).
(a) The entropy change of the refrigerant (ΔS) = Q/T
Q = 180 kJ
T = -15.64 + 273.15 = 257.51 K
ΔS = Q/T
= 180/257.51 = 0.699 kJ/K
(b) The entropy change (ΔS) of the cooled space (ΔS
) = -Q/T
Q = -180 kJ
T = -5 + 273.15 = 268.15 K
ΔS = Q/T
= -180/268.15 = -0.671 kJ/K
(c) The total entropy change for this process (ΔS) = ΔS
+ ΔS
= 0.699 - 0.671 = 0.028 kJ/K