Alcohols have higher boiling points than do ethers and alkanes of similar molar masses.
<h3>What is meant by Boiling Point ?</h3>
The temperature at which the vapor pressure of a liquid equals the pressure surrounding the liquid is called the boiling point of the substance.
Approximately boiling point of Ethanol: 78.4°C, ether: 34.6°C, ethanes: 68°C
The reason for the difference is that it takes more energy to separate alcohol molecules then it does to separate alkane molecules and ether molecules.
The OH group in alcohol allows molecules to engage in hydrogen bonding.
Therefore Boiling point are in the order ;
Alcohol > alkane > ethers
To know more about Boiling Point
brainly.com/question/2153588
#SPJ1
Answer:
If there was only 10 hours in a day, including the night time it would be complexly new environment. Say you are up for 7 hours and sleep 3, in those 7 hours you would go to work for probably 4 hours, come home and do stuff for 3 then go to bed and do it all over again.
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)
Answer:
LOD = 0,0177
LOQ = 0,0345
Explanation:
Detection limit (LOD) is defined as the lowest signal which, with a stated probability, can be distinguished from a suitable blank signal. In the same way, quantification limit (LOQ) is defined as the lowest analyte concentration that can be quantitatively detected with a stated accuracy and precision.
There are many formulas but the most used are:
LOD = X + 3σ
LOQ = X + 10σ
Where X is average and σ is standard desvation
For the blanks readings the average X is 0,0105 and σ is 0,0024
Thus:
<em>LOD = 0,0177</em>
<em>LOQ = 0,0345</em>
I hope it helps!
Explanation:
The volumetric flow rate of water will be as follows.
q = 
= 0.0378 
Diameter = 
= 0.2032 m
Relation between area and diameter is as follows.
A = 
= 
= 0.785 x 0.2032 x 0.2032
= 0.0324 
Also, q = A × V
or, V = 
= 
= 1.166 m/s
As, viscosity of water = 1 cP = 
Pa-s
Density of water = 1000 
Therefore, we will calculate Reynolds number as follows.
Reynolds number = 
= 
= 236931.2
Hence, the flow will be turbulent in nature.
Thus, we can conclude that the Reynolds number is 236931.2 and flow is turbulent.
