The heat released by the water when it cools down by a temperature difference AT
is Q = mC,AT
where
m=432 g is the mass of the water
C, = 4.18J/gºC
is the specific heat capacity of water
AT = 71°C -18°C = 530
is the decrease of temperature of the water
Plugging the numbers into the equation, we find
Q = (4329)(4.18J/9°C)(53°C) = 9.57. 104J
and this is the amount of heat released by the water.
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M
No
Of course not cause it will only locate where the ocean is , it will not show the age.
Answer: Option (A) is the correct answer.
Explanation:
Force acting on a dam is as follows.
F =
.......... (1)
Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.
F' = 
F' =
........... (2)
Now, dividing equation (1) by equation (2) as follows.
=
Cancelling the common terms we get the following.
=
4F = F'
Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.
Sodium will have a larger radius. Look up the atomic radius Trend