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The reaction that the graph represents is
A. Exothermic because Hrxn=-167 kJTo calculate Hrxn we apply the following equation:
Looking at the graph, and at the result of the calculations, we can see that the enthalpy of the products is
lower than the enthalpy of the reagents, because the sign is negative. That means that the reaction
releases energy in the form of heat and that the reaction is
exothermic.
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<h3>Answer:</h3><h3>1865.5g</h3><h3>Explanation:</h3><h3 /><h2> first the chemical formular for ammonium hydroxide is NH4OH</h2><h3>its molarmass is given as N=14H=1O=16 </h3><h3> so we have 14 +1(2) +16+1 =35</h3><h2>also no of moles = mass / molarmass</h2><h3> we have 5.33×10 = mass/35 </h3><h2>therefore mass = 35 ×5.33×10 = 1865.5g</h2>
Answer:
0.7457 g is the mass of the helium gas.
Explanation:
Given:
Pressure = 3.04 atm
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25.0 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
3.04 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K
<u>⇒n = 0.1863 moles</u>
Molar mass of helium = 4.0026 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>0.7457 g is the mass of the helium gas. </u>
Explanation:
Since pressure remained constant, we can eliminate P from the equation
Doing some algebra and converting temperature to Kevin by adding 273, you should obtain the same result.
Answer:
Q= 245 =2.5 * 10^2
Explanation:
ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK
R= 8,314 J/molK, T=298K
ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol
ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ
→ 5.5 = LnQ → Q= 245 =2.5 * 10^2
