Answer:
Some metals can be extracted from compounds by heating with carbon atom because they are less reactive than carbon and some metals cannot be extracted because they are more reactive than carbon atom.
Explanation: If the metal is less reactive than carbon atom so the carbon atoms make bond with oxide or other atom present with metal and the metal is free from that oxide or that element. But if the reactivity of metal is higher than carbon is unable to remove the oxide or element.
Answer:
The answer to your question is: letter A
Explanation:
If an atom lose 3 electrons its charge will be positive, it will be +3
This assumption is not valid because, there are some elements which exist in two or more forms; they have the same atomic number but differ in their mass number, which meas that they possess different number of neutrons. These type of element are called isotopes. Isotope have the same atomic number and similar physical and chemical properties but they have different number of neutrons and therefore possess different masses.
Answer:
Approximately 
, assuming that this acid is monoprotic.
Explanation:
Assume that this acid is monoprotic. Let 
denote this acid.
.
Initial concentration of without any dissociation:
![[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D%20%3D%200.730%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D)
.
After 
of that was dissociated, the concentration of both 
and 
(conjugate base of this acid) would become:
.
Concentration of in the solution after dissociation:
.
Let ![[{\rm HA}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20HA%7D%5D)
, ![[{\rm H}^{+}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D)
, and ![[{\rm A}^{-}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D)
denote the concentration (in 
or 
) of the corresponding species at equilibrium. Calculate the acid dissociation constant 
for , under the assumption that this acid is monoprotic:
![\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DK_%7B%5Crm%20a%7D%20%26%3D%20%5Cfrac%7B%5B%7B%5Crm%20H%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20A%7D%5E%7B-%7D%5D%7D%7B%5B%7B%5Crm%20HA%7D%5D%7D%20%5C%5C%20%26%3D%20%5Cfrac%7B%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%20%5Ctimes%20%280.09125%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%29%7D%7B0.63875%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D%7D%5C%5C%5B0.5em%5D%26%5Capprox%201.30%20%5Ctimes%2010%5E%7B-2%7D%20%5Cend%7Baligned%7D)
.
Answer:
108.5
Explanation:
half of 150 is 75 and half of that is 32.5. Add that to 75 and you'll get your answer
