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Contact [7]
3 years ago
11

ANSWER ASAP AND I WILL GIVE BRAINLYEST!!!!

Chemistry
1 answer:
den301095 [7]3 years ago
8 0

Answer:

C

Explanation:

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The greater the mass of an object
andreyandreev [35.5K]
The greater the mass of an object, the greater<span> its gravitational force.
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Plz chose as brainliest answer plz :)  <#



6 0
3 years ago
Quickly Please!!!!!!
FrozenT [24]

Answer:

n=4.1mol

Explanation:

Hello,

In this case, we can use the ideal gas equation:

PV=nRT

So we know the temperature, pressure and volume, therefore we can easily compute the required moles as shown below:

n=\frac{PV}{RT}=\frac{2.0atm*101L}{0.082\frac{atm*L}{mol*K}*600K}  \\\\n=4.1mol

Best regards.

5 0
3 years ago
Read 2 more answers
Since the number of atoms in a substance is so large , a unit to count them was created. This unit is the number of atoms in 12
larisa86 [58]

The number of atoms in one mole of any substance is measured by Avogadro's number. The value of Avogadro's number is 6.023 x 10 ^23. It is named after scientist Avogadro who proposed this number. 12 grams of carbon-12 represents 1 mole of carbon-12. For this reason, the number of atoms present in 1 mole of any substance is 6.023 x 10 ^23. Therefore, the number of atoms present in 1 mole carbon-12 is 6.023 x 10^23.


(Answer) This unit is the number of atoms in 12 grams of carbon-12 and known as Avogadro's number.

7 0
3 years ago
Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall bala
Nataly [62]

Answer:

E° = 1.24 V

Explanation:

Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)

According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:

Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an

E° = 0.80 V - (-0.44 V) = 1.24 V

6 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
3 years ago
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