<span>Carbon must share 4 electrons total with the the two Oxygen atoms in order to fill it's outer electron shell, and each Oxygen atom must share 2 electrons with the Carbon atom to fill their electron shells. Therefore, in total 8 electrons must be shared (4+2+2=8)</span>
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
The values of the coefficients would be 4, 5, 4, and 6 respectively.
<h3>Balancing chemical equations</h3>
The equation of the reaction can be represented by the following chemical equation:
ammonia (g) + oxygen (g) ---> nitrogen monoxide (g) + water (g)
+
--->
+ 
Thus, the coefficient of ammonia will be 4, that of oxygen will be 5, that of nitrogen monoxide will be 4, and that of water will be 6.
More on balancing chemical equations can be found here: brainly.com/question/15052184
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Bonds between carbon and oxygen are more polar than bonds between sulfur and oxygen. nevertheless, sulfur dioxide (SO₂) exhibits a dipole moment while carbon dioxide (CO₂) does not because of the difference in their shape, CO₂ is having linear geometry thus exhibit zero dipole moment while SO₂ is having bent shape thus exhibit dipole moment. So, despite the fact that bonds between carbon and oxygen are more polar than bonds between sulfur and oxygen. nevertheless, sulfur dioxide (SO₂) exhibits a dipole moment while carbon dioxide (CO₂) does not.
Here is the complete question.
Benzalkonium Chloride Solution ------------> 250ml
Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.
Sig: Dilute 10ml to a liter and apply to affected area twice daily
How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?
(A) 1700 mL
(B) 29.4 mL
(C) 17 mL
(D) 294 mL
Answer:
(B) 29.4 mL
Explanation:
1 L = 1000 mL
1:200 solution implies the
in 200 mL solution.
200 mL of solution = 1g of Benzalkonium chloride
1000 mL will be 
200mL × 1g = 1000 mL × x(g)
x(g) = 
x(g) = 0.2 g
That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.
∴ 
y(g) = 
y(g) = 5g of benzalkonium chloride.
Now, at 17%
concentrate contains 17g/100ml:
∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;
= 
z(mL) = 
z(mL) = 29.41176 mL
≅ 29.4 mL
Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride