Answer:
C: Carbon
Explanation:
Hydrogen, oxygen, and carbon are the most important elements in organic compounds.
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.
The question is incomplete, the complete question is;
Which statement describes a difference between electromagnetic and mechanical waves?
A. Mechanical waves cannot be longitudinal, but electromagnetic waves can.
B. Electromagnetic waves cannot move particles, but mechanical waves can.
C. Electromagnetic waves do not require a medium, but mechanical waves do.
D. Mechanical waves do not transfer energy, but electromagnetic waves do.
Answer:
Electromagnetic waves do not require a medium, but mechanical waves do.
Explanation:
A wave is defined as a disturbance along a medium which transfers energy. Waves may be classified as mechanical waves or electromagnetic waves based on their medium of propagation.
A mechanical wave requires a material medium for propagation. An example of a mechanical wave is sound waves. Sound waves are propagated in air.
Electromagnetic waves do not require a material medium for propagation. They can travel through space. An example of electromagnetic waves is light waves.
Answer:
a) pH = 13.176
b) pH = 13
c) pH = 12.574
d) pH = 7.0
e) pH = 1.46
f) pH = 1.21
Explanation:
HBr + NaOH ↔ NaBr + H2O
∴ equivalent point:
⇒ mol acid = mol base
⇒ (Va)*(0.150mol/L) = (0.025L)*(0.150mol/L)
⇒ Va = 0.025 L
a) before addition acid:
⇒ <em>C </em>NaOH = 0.150 M
⇒ [ OH- ] = 0.150 M
⇒ pOH = - Log ( 0.150 )
⇒ pOH = 0.824
⇒ pH = 14 - pOH
⇒ pH = 13.176
b) after addition 5mL HBr:
⇒ <em>C </em>NaOH = (( 0.025)*(0.150) - (0.005)*(0.150)) / (0.025 + 0.005) = 0.1 M
⇒ <em>C </em>HBr = (0.005)*(0.150) / ( 0.03 ) = 0.025 M
⇒ [ OH- ] = 0.1 M
⇒ pOH = 1
⇒ pH = 13
c) after addition 15mL HBr:
⇒ <em>C </em>NaOH = ((0.025)*(0.150) - (0.015)*(0.150 ))/(0.04) = 0.0375 M
⇒ <em>C </em>HBr = ((0.015)*(0.150))/(0.04) = 0.0563 M
⇒ [ OH- ] = 0.0375 M
⇒ pOH = 1.426
⇒ pH = 12.574
d) after addition 25mL HBr:
equivalent point:
⇒ [ OH- ] = [ H3O+ ]
⇒ Kw = 1 E-14 = [ H3O+ ] * [ OH- ] = [ H3O+ ]²
⇒ [ H3O+ ] = 1 E-7
⇒ pH = 7.0
d) after addition 40mL HBr:
⇒ <em>C</em> HBr = ((0.04)*(0.150) - (0.025)*(0.150)) / (0.04 + 0.025) = 0.035 M
⇒ [ H3O+ ] = 0.035 M
⇒ pH = 1.46
d) after addition 60mL HBr:
⇒ <em>C</em> HBr = ((0.06)*(0.150) - (0.025)*(0.150)) / (0.06+0.025) = 0.062 M
⇒ [ H3O+ ] = 0.062 M
⇒ pH = 1.21
