Answer:
Alkenes are those organic compound which have double carbon bond
Formula of but-2-ene = C4H8
Given:
128g sample of titanium
2808J of heat energy
specific heat of titanium is 0.523 J/ g °C.
Required:
Change in temperature
Solution:
This can be solved
through the equation H = mCpT
where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.
Plugging in the
values into the equation
H = mCpT
2808J = (128g) (0.523
J /g °C) T
T
= 41.9 °C
The density of the carbon tetrachloride is 1.59 g/mL.
Mass of CCl₄ = 703.55 g – 345.8 g = 357.75 g
Density = mass/volume = 357.75 g/225 mL = 1.59 g/mL
Answer:
The correct answer is
C. 6 Protons
Explanation:
Carbon is a nonmetallic element that is available in both organic and inorganic compounds.
Carbon belongs to group 14 elements in the periodic table,carbon is a chemical element with the symbol C and atomic number 6, it can form long chains with its own atoms, a feature called catenation.
Two allotropes of carbon available are diamonds and graphites, which have different crystalline structures
The physical properties of carbon vary widely with the allotropic form.
examples are.
- Graphite, diamonds and coal are all nearly pure forms of carbon
- Diamond is highly transparent. Graphite is opaque and black
- Diamond is one of the hardest substances known to man. Graphite is soft and often used as the "lead" in lead pencils
- Diamond has a very low electrical conductivity. Graphite is a very good conductor
- Very brittle, and cannot be rolled into wires or pounded into sheets
Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681
Explanation: (a) pH of the buffer solution is calculated using Handerson equation:
pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.
pH = 4.76 + 0.014
pH = 4.774
(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.
Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.
moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88
moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99
Let's again plug in the values in the Handerson equation:
pH = 4.76 + 0.051
pH = 4.811
(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).
so, new moles of acetic acid = 0.92 + 0.100 = 1.02
new moles of sodium acetate = 0.95 - 0.100 = 0.85
Let's plug in the values in the equation:
pH = 4.76 - 0.079
pH = 4.681
