Answer:
ΔHrxn = - 1534.3 J
Explanation:
Given the assumptions and the formula for the change in enthalpy:
ΔHrxn = m x C x ΔT, where
m is the mass of solution given 135.4 g
C is the heat capacity 4.2 J/g .K and,
ΔT is the change in temperature
we have ,
T₁ = ( 18.1 + 273) K = 291.1 K
T₂ = ( 15.4 +273) K = 288.4 K
ΔHrxn = 135.3 g x 4.2 J/gK x ( 288.4 -291.1 ) K = - 1534.3 J
After verifying our result has the correct unit, the answer is -1534.3 Joules, and the negative sign tells us it is an endothermic reaction decreasing the final temperature.
The gas is confined in 3.0 L container ( rigid container) ⇒ the volume remains constant when the temperature is increased from from 27oC to 77oC and therefore V1=V2 .
<h2>
Hope it helps you please mark as brainlist</h2>
Given:
35.0 mL of acid with an unknown concentration
24.6 mL of 0.432 M base
Required:
Concentration of the acid
Solution:
M1V1 = M2V2
M1 (35.0 mL of acid)
= (0.432 M base) (24.6 mL
of base)
V1 = (0.432
M base) (24.6 mL of base) /
(35.0 mL of acid)
M1 = 0.304 M of acid
<span>0.38
You first calculate the total moles by dividing the grams by molecular weight:
45 g N2 / 28.02 g/mol = 1.6 mol N2
40 g Ar / 39.95 g/mol = 1.0 mol
Then you divide the moles of Ar by the total number of moles:
1.0 / (1.6 + 1.0) = 0.38 mol fraction</span>
₉₂U²³⁵ + ₀n¹ → ₅₄Xe¹⁴⁰ + ₃₈Sr⁹⁴ + 2 ₀n¹
Mass of reactants = 235.04393 + 1.008665 = 236.052595 amu
Mass of products = 139.92144 + 93.91523 + 2* (1.008665) = 235.854000 amu
Mass defect Δ m = 236.052595 - 235.854000 = 0.198 amu
Reaction energy released Q = Δ m * 931.5
= 0.198 * 931.5 = 185 MeV
