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Eddi Din [679]
2 years ago
6

How is a double displacement reaction completed and balanced

Chemistry
1 answer:
Kisachek [45]2 years ago
7 0

A double displacement reaction is a type of reaction in which two reactants exchange ions to form two new compounds. Many double displacement reactions occur between ionic compounds that are dissolved in water. A double replacement reaction is represented by the general equation.

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Calculate the % composition of the unknown liquid using your most precise result. It is a mixture of ethanol (D[ETOH] = 0.7890 g
Rufina [12.5K]

% composition of ethanol = 34.51%

% composition of water  = 65.49%

<h3>What is density?</h3>

A material's density is defined as its mass per unit volume.

Given data:

The density of ethanol = 0.7890 g/mL

The density of water = 0.9982 g/mL

The density of mixture = 0.926 g/mL

Let the % composition of ethanol = x

Let the % composition of water = 100-x

Now density of the mixture

\frac{Mass}{Volume}

Mass = \frac{percent  \;of  \;ethanol  \;X  \;density  \;of  \;ethanol  \;+  \\ \;percent  \;of  \;water X  \;density  \;of  \;water}{100}

0.926 = \frac{x X  0.7890 g/mL  \;+  (100-x) X  0.9982 g/mL}{100}

x= 34.51 %

Hence,

% composition of ethanol = 34.51%

% composition of water = 65.49%

Learn more about the density here:

brainly.com/question/952755

#SPJ1

8 0
1 year ago
Correct forms of the equation of Charles’s law is (are)
xxTIMURxx [149]

According to Charles' Law the volume of an ideal gas is directly proportional to its absolute temperature in Kelvin keeping the pressure constant.

V∝ T, P  is constant  

where V, T and P are volume, temperature and pressure

\frac{V1}{T1 } = \frac{V2}{T2}

where V₁, T₁, V₂ and T₂ are initial volume, initial temperature, final volume and final temperature.  

8 0
3 years ago
In glycolysis, if glucose is labeled at the carbon 6 position (see page 1 for numbering of carbons in glucose) A) the carbon wit
Oliga [24]

Answer:

D) the carbon with the low-energy phosphate on it in 1,3 BPG is labeled.

Explanation:

Glycolysis has 2 phase (1) preparatory phase (2) pay-off phase.

<u>(1) Preparatory phase</u>

During preparatory phase glucose is converted into fructose-1,6-bisphosphate. Till this time the carbon numbering remains the same i.e. if we will label carbon at 6th position of glucose, its position will remian the same in fructose-1,6-bisphosphate that means the labeled carbon will still remain at 6th position.

When fructose-1,6-bisphosphate is further catalyzed with the help of enzyme aldolase it is cleaved into two 3 carbon intermediates which are glyceraldehyde 3-phosphate (GAP) and dihyroxyacetone  phosphate (DHAP).  In this conversion, the first three carbons of fructose-1,6-bisphosphate become carbons of DHAP while the last three carbons of fructose-1,6-bisphosphate will become carbons of GAP. It simply means that GAP will acquire the last carbon of fructose-1,6-bisphosphate which is labeled. Now the last carbon of GAP which has phosphate will be labeled.  

<u>(2) Pay-off phase</u>

During this phase, GAP is dehydrogenated into 1,3-bisphosphoglycerate (BPG) with the help of enzyme glyceraldehyde 3-phosphate dehydrogenase. This oxidation is coupled to phosphorylation of C1 of GAP and this is the reason why 1,3-bisphosphoglycerate has phosphates at 2 positions i.e. at position 1 in which phosphate is newly added and position 3rd which already had labeled carbon.

It is pertinent to mention here that<u> BPG has a mixed anhydride and the bond at C1 is a very high energy bond.</u> In the next step, this high energy bond is hydrolyzed into a carboxylic acid with the help of enzyme phosphoglycerate kinase and the final product is 3-phosphoglycerate. Hence, the carbon with low energy phosphate i.e. the carbon at 3rd position remains labeled.

3 0
2 years ago
Hydrogen peroxide can decompose to water and oxygen by the following reaction
mestny [16]
To begin calculating, there is one thing you need to remember :1 mole of H2O2=34.0148 g
Then we have 5.00g of H2O2=5/34.0148=0.146995
As you know decomposition of 2moles now has prodused <span>196kj
So, </span><span>q is made due </span>0.146995 moles of H2O2=(-196/2)*0.146995=-14.40551Kj
I'm sure it will help.
6 0
2 years ago
Justify air is a mixture not a compound
Andreas93 [3]
A compound has to be chemically bonded, however, air is not chemically bonded.
This can be proven by freezing air. By freezing air, it yields different liquids at different temperature. Liquid nitrogen has a different boiling point than liquid oxygen.
If air was a compound, they would all have a single boiling point and a single freezing point.


Hope this helps :)
4 0
3 years ago
Read 2 more answers
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