Answer:
If the temperature increases the molecular movement as well, and if it increases the same it will happen with the molecular movement.
Pressure, volume and temperature are three factors that are closely related since they increase the temperature, the pressure usually decreases due to the dispersion of the molecules that can be generated, so the volume also increases.
If the temperature drops, the material becomes denser, its molecules do not collide with each other, their volume and pressure increases.
Explanation:
The pressure is related to the molecular density and the movement that these molecules have.
The movement is regulated by temperature, since if it increases, the friction and collision of the molecules also.
On the other hand, the higher the volume, the less pressure there will be on the molecules, since they are more dispersed among themselves.
(in the opposite case that the volume decreases, the pressure increases)
The subatomic particle that identifies the atom is the number of protons. This is what distinguishes an element that is is flammmable, hydrogen to one that is essential component in water, oxygen.
ANSWER
The correct answer is A
EXPLANATION
Plants manufacture their on food by the process of photosynthesis. During this process, plants trap radiant energy from the sun by the help of chlorophyll in the leaves.
Radiant energy with other raw materials such as water, carbon dioxide and mineral salts is converted to food (in the form of starch) which contains chemical energy.
Answer:
0.3808
Explanation:
number of moles,n=Conc.XVol.
hence 0.85X0.448
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
