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igomit [66]
2 years ago
12

Determine the mass of 10.55 x 1029 molecules of cadmium chloride, CdCl,().

Chemistry
1 answer:
VashaNatasha [74]2 years ago
3 0

Answer:

Moles= 10.55*10^29ǜ.02*10^23

=1.75*10^6 moles

Moles= mass/molar mass

1.75*10^6=mass/ 183

Mass= 320*10^6 g

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Please indicate which of the compounds will yield postive results for the permanganate, ninhydrin, and ceric ammonium nitrate te
maks197457 [2]

The compounds will yield Positive results for the permanganate, ninhydrin, and ceric ammonium nitrate tests  are

permanganate test= ethene,

ninhydrin=aspasyic acid Phenylalanine

ceric ammonium nitrate test= methanol

<h3>What is a compound?</h3>

Generally, A compound is simply defined as a substance made up of two or more different chemical elements held together by chemical bonds that are difficult to break.

In conclusion, For the permanganate test= ethene,

ninhydrin=aspasyic acid Phenylalanine

ceric ammonium nitrate test= methanol

Read more about Compound

brainly.com/question/704297

3 0
1 year ago
Question 2(Multiple Choice Worth 4 points) (01.01 MC) Dominic and Eva are using the same type of stopwatches to measure the time
LekaFEV [45]
The results of Dominic and Eva's experiment is unreliable and can lead to a pseudoscientific claim primarily because they did not repeat their tests multiple times. Although, they used the same type of stopwatch in recording the time it takes for the chemical reaction to occur, they have different reactions times. Thus, it would have been better if they conducted several trials then obtained the average of their results.    
6 0
2 years ago
Read 2 more answers
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
PLEASE ANSWER I WILL GIVE YOU BRAINIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
blondinia [14]

Answer:

Erosion

Explanation:

Erosion is not a cause of metamorphism

The causes are mostly temperature, pressure, heat from the metamorphic rock

5 0
2 years ago
Read 2 more answers
The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of
timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0
3 years ago
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