Answer:
Liquid B because of its higher vapor pressure due to the fact that evaporation rate is directly proportional to vapor pressure
Explanation:
The vapor pressure of liquid at equilibrium is a function to the liquid's rate of evaporation. The evaporation rate and hence the vapor pressure is a measure of the propensity of the particles of the liquid to leave the surface of the liquid and exist as vapor directly above the liquid. As high evaporation rate leads to high vapor pressure, a liquid with a higher vapor pressure will evaporate faster than one with a lower vapor pressure at the same temperature and pressure.
Therefore, liquid B with a vapor pressure of 18.04 kPa at 40° C will evaporate faster than liquid A with a lower vapor pressure of 7.37 kPa at the same 40°C.
Answer:
![RCOO^{-}\\RNH_{2}\\H_{2}PO_{4}^{-}\\HCO_{3-}](https://tex.z-dn.net/?f=RCOO%5E%7B-%7D%5C%5CRNH_%7B2%7D%5C%5CH_%7B2%7DPO_%7B4%7D%5E%7B-%7D%5C%5CHCO_%7B3-%7D)
Explanation:
A conjugate base is formed when an acid in reaction with a base lose a proton H+.
For example if RCOOH that is an acid, reacts with water that is a base, the RCOOH loses an H+ and it becomes
that is the conjugate base; and the water gains the H+ and become the conjugate acid, that is:
* ![RCOOH, RCOO-](https://tex.z-dn.net/?f=RCOOH%2C%20RCOO-)
![RCOOH+H_{2}O=H_{3}O^{+}+RCOO^{-}](https://tex.z-dn.net/?f=RCOOH%2BH_%7B2%7DO%3DH_%7B3%7DO%5E%7B%2B%7D%2BRCOO%5E%7B-%7D)
The RCOO- is the conjugate base
* ![RNH_{2}, RNH_{3}^{+}](https://tex.z-dn.net/?f=RNH_%7B2%7D%2C%20RNH_%7B3%7D%5E%7B%2B%7D)
![RNH_{3}^{+}+H_{2}O=H_{3}O^{+}+RNH_{2}](https://tex.z-dn.net/?f=RNH_%7B3%7D%5E%7B%2B%7D%2BH_%7B2%7DO%3DH_%7B3%7DO%5E%7B%2B%7D%2BRNH_%7B2%7D)
In this case the
is the conjugate base
* ![H_{3}PO_{4},H_{2}PO_{4}^{-}](https://tex.z-dn.net/?f=H_%7B3%7DPO_%7B4%7D%2CH_%7B2%7DPO_%7B4%7D%5E%7B-%7D)
![H_{3}PO_{4}+H_{2}O=H_{3}O^{+}+H_{2}PO_{4}^{-}](https://tex.z-dn.net/?f=H_%7B3%7DPO_%7B4%7D%2BH_%7B2%7DO%3DH_%7B3%7DO%5E%7B%2B%7D%2BH_%7B2%7DPO_%7B4%7D%5E%7B-%7D)
The
is the conjugate base
* ![H_{2}CO_{3},HCO_{3}^{-}](https://tex.z-dn.net/?f=H_%7B2%7DCO_%7B3%7D%2CHCO_%7B3%7D%5E%7B-%7D)
![H_{2}CO_{3}+H_{2}O=H_{3}O^{+}+HCO_{3}^{-}](https://tex.z-dn.net/?f=H_%7B2%7DCO_%7B3%7D%2BH_%7B2%7DO%3DH_%7B3%7DO%5E%7B%2B%7D%2BHCO_%7B3%7D%5E%7B-%7D)
The
is the conjugate base.
Answer:
The amount present is 36g
Explanation:
Here, we are interested in calculating the amount of zinc in the bronze alloy since it is 18% of the total mass
What we do here is to calculate 18% of the total 200g
Mathematically, that would be 18/100 * 200 = 18 * 2 = 36g