I believe the answer would be C
The the compound of interest i.e. XCl4, since there are 4 Cl atoms bonded to X. This signifies that the valency of X is 4.
There atomic number of C is 6. It's electronic configuration is giving by 1s2 2s2 2p2. Thus, there are 4 electrons in valence shell of C. This signifies that valency of C is 4. Hence the compound present in present case is CCl4.
Hope this helps :)
Answer:
.997 atm
Explanation:
1. Find the combined gas law formula...
(P1V1/T1 = P2V2/T2)
2. Find our numbers...
P1= .982 atm
P2= ? (trying to find)
V1= 2 L
V2= 1.8 L
T1= 22 C = 295 K
T2= -3 C = 270 K
- Note: always use Kelvin. To find Kelving add 273 to ___C.
3. Rearrange formula to fit problem...
(P2=P1V1T2/V2T1)
4. Fill in our values...
P2= .982 atm x 2 L x 270 K / 1.8 L x 295 K
5. Do the math and your answer should be...
.997 atm
- If you need more help or still do not understand please let me know and I would be glad to help!
Answer:
a. withdraws electrons inductively
b. donates electrons by hyperconjugation
c. donates electrons by resonance
d. withdraws electrons inductively
Explanation:
a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:
F > Cl > Br>I
b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.
c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.
d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.
The answer is
<span>The density (D) is quotient of mass (m) and
volume (V):
</span>
The unit is g/cm³
It is given:
m = 1.62 kg = 1620 g
V = 205 mL = 205 cm³
D = ?
Thus:
The density of the goblet is 7.90 g/cm³.
Answer:
[HClO₄] = 11.7M
Explanation:
First of all we need to know, that a weight percent represents, the mass of solute in 100 g of solution.
Let's convert the mass to moles → 70.5 g . 1mol/100.45 g = 0.702 moles
Now we can apply the density to calculate the volume.
Density always refers to solution → Solution density = Solution mass / Solution volume
1.67 g/mL = 100 g / Solution volume
Solution volume = 100 g / 1.67 g/mL → 59.8 mL
To determine molarity (mol/L) we must convert the mL to L
59.8 mL . 1L/1000mL = 0.0598 L
Molarity → Moles of solute in 1L of solution → 0.702 mol / 0.0598 L = 11.7M
