Answer:
2.8
Explanation:
First, we will calculate the molarity of the acetylsalicylic acid solution.
M = mass of solute (g) / molar mass of solute × volume of solution (L)
M = 0.327 g / 180.158 g/mol × 0.237 L
M = 7.66 × 10⁻³ M
For a weak acid such as acetylsalicylic acid, we can find the concentration of H⁺ using the following expression.
[H⁺] = √(Ca × Ka)
where,
Ca: concentration of the acid
Ka: acid dissociation constant
[H⁺] = √(7.66 × 10⁻³ × 3.3 × 10⁻⁴)
[H⁺] = 1.6 × 10⁻³ M
The pH is:
pH = -log [H⁺]
pH = -log 1.6 × 10⁻³ = 2.8
The density would be the same for the whole bar as well as one half of the bar. Density is a identity I believe, by this I mean that it stays the same no matter how little or how much of the same substance you have. Since density = mass / volume, half the bar has half of the weight as well as half of the volume of the whole bar, making the density the same.
For example, a block weighs 10 grams and has a volume of 5 ml. the density would be d = 10/5 or, d = 2g/ml
Half of the block weighs 5 grams and has a volume of 2.5 ml. The density is d = 5/2.5, or, d = 2 g/ml.
See, although there are different amounts of the same substance, their density is the same.
Explanation:
Acceleration (a) = 600 m/s²
Mass (m) = 300 kg
Force (F) = ?
We know
F = m * a
= 300 * 600
=180000 Newton
The force acting on the object is 18000 Newton.
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where 
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So, for the given reaction is -99.4 J/K
Answer:
589.038 m/s
Explanation:
i dont know if did this right tho
