Answer:
The final temperature of the solution is 44.8 °C
Explanation:
assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:
Q dis + Q sol = 0
Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C) is q dis= -83.3 KJ/mol . And the molecular weight is
M = 1*40 g/mol + 2* 35.45 g/mol = 110.9 g/mol
Q dis = q dis * n = q dis * m/M = -83.3 KJ/mol * 13.1 g/110.9 gr/mol = -9.84 KJ
Qdis= -9.84 KJ
Also Qsol = ms * Cs * (T - Ti)
therefore
ms * Cs * (T - Ti) + Qdis = 0
T= Ti - Qdis * (ms * Cs )^-1 =24.7 °C - (-9.84 KJ/mol)/[(104 g + 13.1 g)* 4.18 J/g°C] *1000 J/KJ
T= 44.8 °C
Answer:
1.88 M
Explanation:
The following data were obtained from the question:
Mole of NaCl = 0.47 mole
Volume of solution = 0.25L
Molarity =?
Molarity is defined as the mole of solute per unit litre of the solution. It can represented mathematically as:
Molarity = mole /Volume
Using the above formula, the molarity of the salt water solution can be obtained as follow:
Molarity = 0.47/0.25
Molarity = 1.88 M
Answer is: elements in group 1 will lose electrons to obtain a noble gas structure. They will lose 1 electron.
For example ₃Li 1s²2s¹ will lose one electron from 2s oribtal to obtain helium structure ₂He 1s².
Or sodium ₁₁Na 1s²2s²2p⁶3s¹ will also lose one electron to obtain neon structure ₁₀Ne 1s²2s²2p⁶.
<span>Xe = VIII = 8 valence electrons
F = VII = 4 (7 ve) = 28 valence electrons</span>
total ve = 8 + 28 = 36 ve
<span>36 - 4(2) = 28 ve
(there are 2 electrons in each bond x 4 bonds)</span>
<span>28 - 4(6) = 4
(We assign the remaining electrons to F atoms)</span>
<span>4 - 2(2) = 0
(Therefore 4 electrons left => we have 2 lone pairs)</span>
The steric number = No. of
σ bonds + #lone pairs
= 4 σ bonds + 2 lone pairs
= 6 => d²sp³ (6 hybrid orbitals)
<span>4 bonds + 2 lone pairs
=> square planar</span>
Answer:
option A is correct answer
