Because the molecular solutions have low melting and boiling points, and do not conduct electricity.
Hey there!
No of hybrid orbitals , H = ( V +S - C + A ) / 2
Where H = no . of hybrid orbitals
V = Valence of the central atom = 5
S = No . of single valency atoms = 4
C = No . of cations = 1
A = No . of anions = 0
For PCl4 +
Plug the values we get H = ( 5+4-1+0) / 2
H = 4 ---> sp3 hybridization
sp3 hybrid orbitals are used by phosphorous in the PCl4+ cations
Answer C
Hope that helps!
Harmonic melody and something else hold on...checking
Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place
Answer:
37.1°C.
Explanation:
- Firstly, we need to calculate the amount of heat (Q) released through this reaction:
<em>∵ ΔHsoln = Q/n</em>
no. of moles (n) of NaOH = mass/molar mass = (2.5 g)/(40 g/mol) = 0.0625 mol.
<em>The negative sign of ΔHsoln indicates that the reaction is exothermic.</em>
∴ Q = (n)(ΔHsoln) = (0.0625 mol)(44.51 kJ/mol) = 2.78 kJ.
Q = m.c.ΔT,
where, Q is the amount of heat released to water (Q = 2781.87 J).
m is the mass of water (m = 55.0 g, suppose density of water = 1.0 g/mL).
c is the specific heat capacity of water (c = 4.18 J/g.°C).
ΔT is the difference in T (ΔT = final temperature - initial temperature = final temperature - 25°C).
∴ (2781.87 J) = (55.0 g)(4.18 J/g.°C)(final temperature - 25°C)
∴ (final temperature - 25°C) = (2781.87 J)/(55.0 g)(4.18 J/g.°C) = 12.1.
<em>∴ final temperature = 25°C + 12.1 = 37.1°C.</em>