Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
Answer: Option (d) is the correct answer.
Explanation:
In solids, molecules are held together by strong intermolecular forces of attraction. As a result, they are unable to move from their initial position and can only vibrate at their mean position.
Hence, a solid has definite shape and volume. Solids cannot be compressed.
Whereas in plasma, molecules are hot ionized which include positively charged ions and negatively charged electrons. They collide much more rapidly with each other and are widely spreaded out.
Therefore, they occupy the volume of container in which they are placed. Plasma can be compressed.
Thus, we can conclude that substance X is a solid and substance Y is a plasma.
Answer:
false
Explanation:
Atoms of the same element that differ in their numbers of neutrons are called isotopes. Many isotopes occur naturally. ... Different isotopes of an element generally have the same physical and chemical properties because they have the same numbers of protons and elec
Answer:
The elements in Group 2 (beryllium, magnesium, calcium, strontium, barium, and radium) are called the alkaline earth metals (see Figure below). These elements have two valence electrons, both of which reside in the outermost s sublevel. The general electron configuration of all alkaline earth metals is ns
Answer : The activation energy of the reaction is, 
Solution :
The relation between the rate constant the activation energy is,
![\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial rate constant = 
= final rate constant = 
= initial temperature = 
= final temperature = 
R = gas constant = 8.314 kJ/moleK
Ea = activation energy
Now put all the given values in the above formula, we get the activation energy.
![\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7B8.75%5Ctimes%2010%5E%7B-3%7DL%2Fmole%5Ctext%7B%20s%7D%7D%7B4.55%5Ctimes%2010%5E%7B-5%7DL%2Fmole%5Ctext%7B%20s%7D%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20%288.314kJ%2FmoleK%29%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B468K%7D-%5Cfrac%7B1%7D%7B531K%7D%5D)
Therefore, the activation energy of the reaction is,
