Answer:
The mole fraction of ethanol is 0.6. A 10 mL volumetric pipette must be used for to measure the 10 mL of ethanol. The vessel should be clean and purged.
Explanation:
For calculating mole fraction of ethanol, the amount of moles ethanol must be calculated. Using ethanol density (0.778 g/mL), 10 mL of ethanol equals to 7.89 g of ethanol and in turn 0.17 moles of ethanol. The same way for calculate the amount of water moles (ethanol density=0.997 g/mL). 2 mL of water correspond to 0.11. The total moles are: 0.17+0.11=0.28. Mole fraction alcohol is: 0.17/0.28=0.6
Answer:
H2Br + 2KOH ----- K2Br + 2H2O
Answer:
0.0010 mol·L⁻¹s⁻¹
Explanation:
Assume the rate law is
rate = k[A][B]²
If you are comparing two rates,
![\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7Brate%7D_%7B2%7D%7D%7B%5Ctext%7Brate%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7Bk_%7B2%7D%5Ctext%7B%5BA%5D%7D_2%5B%5Ctext%7BB%5D%7D_%7B2%7D%5E%7B2%7D%7D%7Bk_%7B1%7D%5Ctext%7B%5BA%5D%7D_1%5B%5Ctext%7BB%5D%7D_%7B1%7D%5E%7B2%7D%7D%3D%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%5Cright%20%29%20%5Cleft%20%28%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%5Cright%20%29%5E%7B2%7D)
You are cutting each concentration in half, so
![\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7B%5BA%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BA%5D%7D_%7B1%7D%7D%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Ctext%7B%20and%20%7D%5Cdfrac%7B%5Ctext%7B%5BB%5D%7D_%7B2%7D%7D%7B%5Ctext%7B%5BB%5D%7D_%7B1%7D%7D%3D%20%5Cdfrac%7B1%7D%7B2%7D)
Then,

Exothermic reactions release heat the surrounding. So Freezing?
Answer:
Basically an object in motion or at rest will remain that way unless an outside force acts against it.
Explanation:
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This is normally taken as the definition of inertia.