The larger the kinetic energy of the vehicle, the larger the amount of energy will be needed to stop the vehicle, meaning that faster vehicles have a larger stopping distance
<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.
<u>Explanation:</u>
We are given:
39.0 % of sodium in sodium chloride solution
This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution
Mass of sodium given = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)
Applying unitary method:
If 39 grams of sodium metal is present in 100 grams of sodium chloride solution
So, if 0.1 grams of sodium metal will be present in = of sodium chloride solution.
Hence, the mass of sodium chloride solution present is 0.256 grams.
Answer: The frequency of the wave is 0.5 hertz.
Explanation:
= Frequency of the wave
= speed of the light in m/s
= Wavelength of the wave.
Here in question we are given with speed of the infrared light. So, we will replace the value of speed of light(c) from the given value of the speed of the infrared light.
Speed of infrared light = 6 m/s
The frequency of the wave is 0.5 hertz.
Q = M * C *ΔT
Q / <span>ΔT = M
</span>Δf - Δi = 98.4ºC - 62.2ºC = 36.2ºC
<span>
C = 1137 J / 140 * 36.2
C = 1137 / 5068
C = 0.224 J/gºC</span>
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