Answer:
When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since P and V are inversely proportional, a graph of 1/P vs. V is linear.
Explanation:
<h2><u><em>
This is all I know.</em></u></h2><h2><u><em>
Hope it helps~</em></u></h2><h2><u><em>
Good luck with your work!!~</em></u></h2>
Answer:
Explanation:
M(s) → M (g ) + 20.1 kJ --- ( 1 )
X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )
M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )
( 3 ) - 2 x ( 2 ) - ( 1 )
M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s) → M X₄ (g ) - 98.7 kJ - 2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ
0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ
4 X (g ) + M (g ) = M X₄ (g ) - 773.4kJ
heat of formation of M X₄ (g ) is - 773.4 kJ
Bond energy of one M - X bond = 773.4 / 4 = 193.4 kJ / mole
Answer:
The percentage of the block contains 15% after 60 years.
Explanation:
Step 1: Formula for half-life time
To calculate the half-life time we will use the following formula:
At = A0 * 1/2 ^(x/t)
With At = the quantity after a time t
A0 = The quantity at time t = 0 (start)
x = time in this case = 60 years
t= half-life time = 22 years
Step 2: Calculate the percentage after 60 years
In this case: after 60 years the percentage will be
A0= 10 * 1/2 ^(60/22)
A0 = 1.5
A0 / At = 1.5 /10 = 0.15
0.15 *100% = 15 %
The percentage of the block contains 15% after 60 years.
Answer:
Los habitantes del planeta con una atmósfera superior a 5,1 atm de la Tierra, no estarían nadando en ríos de dióxido de carbono líquido
Explanation:
De las tablas de datos termodinámicos, la presión a la que el vapor de dióxido de carbono está en equilibrio con su estado líquido a una temperatura ambiente de 25 ° C es 6,401 kPa, lo que equivale a 63,17296 atm.
Por lo tanto, a una presión de 5.1 de la atmósfera terrestre, el dióxido de carbono es completamente gaseoso y los habitantes del planeta con una presión atmosférica de 5.1 atm de la Tierra todavía observarían solo hidrógeno gaseoso y no estarían nadando en ríos de dióxido de carbono líquido.
<u>Given:</u>
H2(g) + Cl2 (g) → 2HCl (g)
<u>To determine:</u>
The enthalpy of the reaction and whether it is endo or exothermic
<u>Explanation:</u>
Enthalpy of a reaction is given by the difference between the enthalpy of formation of reactants and products
ΔH = ∑nHf (products) - ∑nHf (reactants)
= [2Hf(HCl)] - [Hf(H2) + Hf(Cl2)] = 2 (-92.3) kJ = - 184.6 kJ
Since the reaction enthalpy is negative, the reaction is exothermic
<u>Ans:</u> The enthalpy of reaction is -184. kJ and the reaction is exothermic