Answer: combustion causes a chemical reaction between nitrogen and oxygen in the engine.
Explanation:
Nitrogen oxides are produced in combustion processes, partly from nitrogen compounds in the fuel, but mostly by direct combination of atmospheric oxygen and nitrogen in flames. Nitrogen oxides are produced naturally by lightning, and also, to a small extent, by microbial processes in soils.
Cylinder D
temperature is the measurement of average kinetic energy of a sunstance. since cylinder D has the highest average velocity, it has the highest average kinetic energy among the 4 cylinders, thus cylinder D is the hottest.
Answer:
As a substance melts, and goes from a solid to a liquid state, the kinetic energy of the molecules increases, and the molecules move faster, and they separate further and further away from each other. The intermolecluar forces holding the molecules together become weaker. This is why a liquid can take fill the shape of its container, whereas a solid has a fixed shape.
Explanation:
take your notes man
The statement which is true about the reactivity of element with 1S²2S²2P⁶3S¹ is
it is reactive because it has to lose one electron to have a full outermost energy level.
<u><em>Explanation</em></u>
- <u><em> </em></u>Element with 1S²2S²2P⁶3S¹ electron configuration is a sodium metal.
- sodium has one electron in the outermost energy level.
- for sodium to have a full outermost energy level ( 8 electrons) it loses the 1 electron in 3S¹ to form a positively charged ion. (Na⁺)
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 %