<span>The Lewis structure for CO has 10 valence electrons. For the CO Lewis structure you'll need a triple bond between the Carbon and Oxygen atoms in order to satisfy the octets of each atom while still using the 10 valence electrons available for the CO molecule.</span>
Answer:
406.45mL
Explanation:
The following data were obtained from the question:
V1 = 350mL
P1 = 720mmHg
P2 = 630mmHg
V2 =?
The new volume can be obtain as follows:
P1V1 = P2V2
720 x 350 = 620 x v2
Divide both side by 620
V2 = (720 x 350) /620
V2 = 406.45mL
The new volume of the gas is 406.45mL
Answer:
mol LiCl = 4.83 m
Explanation:
GIven:
Solution of LiCl in water XLiCl = 0.0800
Mol of water in kg = 55.55 mole
Find:
Molality
Computation:
mole fraction = mol LiCl / (mol water + mol LiCl)
0.0800 = mol LiCl / (55.55 mol + mol LiCl)
0.0800 mol LiCl + 4.444 mol = mol LiCl
mol LiCl - 0.0800 mol LiCl = 4.444 mol
0.92 mol LiCl = 4.444 mol
mol LiCl = 4.83 m
Volume of Cl₂(g) is produced at 1.0 atm and 540.°C=4.5×10^4 L
As per the evenly distributed response
2NaCl (g) ----> 2Na(l)+ Cl2(g)
Calculate the amount of Cl2 that was formed as indicated below:
Moles of Cl2 = 31.0 kg of Na x (1000* 1 * 1 / 1*23* 2)
= 673.9 mol
P is equal to 1.0 atm, and T is equal to 813.15 K
when converted to Kelvin by multiplying by a factor of 273.15.
Using Cl2 as an ideal gas, determine the in the following volume:
volume = nRT/P
= 673.9 * 0.0821 * 813.15/ 1
=4.5×10^4 L
As a result, the volume of Cl2 under the given circumstances =4.5×10^4 L
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Answer:
4.90 g
Explanation:
Given that:
volume of t-pentyl alcohol = 5 mL
the standard density of t-pentyl alcohol = 0.805 g/mL
Recall that:
density = mass(in wt) /volume
mass = density × volume
mass = 0.805 g/mL × 5 mL
mass = 4.03 g
Volume of HCl used = 12 mL
The reaction for this equation is shown in the image attached below.
From the reaction,
88.15 g of t-pentyl alcohol reacts with concentrated HCl to yield 106.59 g pf t-pentyl chloride.
4.03 g of t-pentyl alcohol forms,
of t-pentyl chloride.
Therefore,
Theoretical yield of t-pentyl chloride = 4.90 g