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stellarik [79]
3 years ago
14

What type of mixture scatters light and cannot be filtered?

Chemistry
1 answer:
GenaCL600 [577]3 years ago
3 0
Medium sized particles 
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Some glucose produced by gluconeogenesis is stored in the body as glycogen. Order the steps of glycogen synthesis.
Harlamova29_29 [7]

Answer:

e. UDP-glucose pyrophosphorylase catalyzes the reaction of glucose-I-phosphate and UTP to UDP-glucose and PPi

a. Pyrophosphatase converts PPi and water into two Pi

b. Glycogen synthase adds a glucose unit from UDP-glucose to glycogen, producing a larger glycogen molecule and UDP

Explanation:

Glycogen synthesis or glycogenesis is the process of synthesis of glycogen molecules from glucose molecules in living organisms. Glycogen is a polysaccharide storage form of glucose and helps to store excess glucose in the body form use when required by the body.

The synthesis of glycogen involves sugar nucleotides. Sugar nucleotides are compounds in which a sugar molecule is attached to a nucleotide through phosphate ester bond, resulting in the activation of the sugar molecule. The sugar nucleotides then are used as substrates for the polymerization of the monosaccharide sugars into disaccharides, oligosaccharides and polysaccharides.

In the synthesis of glycogen, glucose-6-phosphate from phosphorylation of free glucose by hexokinase is first isomerized to glucose-1-phosphate by phosphoglucomutase.

Glucose-1-phosphate is then converted to UDP-glucose by its reaction with UTP catalyse by UDP-glucose pyrophosphorylase. The reaction is favoured by the rapid hydrolysis of PPi produced to two molecules of inorganic phosphate by the enzyme pyrophosphatase.

Glycogen synthase then adds a glucose unit from UDP-glucose to a growing chain of glycogen, producing a larger glycogen molecule and free UDP.

6 0
3 years ago
Rubbing alcohol evaporates from your hand quickly, leaving a cooling effect on your skin. Because evaporation is an example of a
Ludmilka [50]
The molecules in gas are farther apart and have more room to bounce around than liquid
3 0
3 years ago
Read 2 more answers
Locations Cost of land Cost of equipment Cost of mining and reclamation Time taken to mine the area 1 $100,000 $10,000 $5,000 pe
mario62 [17]

Answer:

Location 2 will cost the least to the company to mine.

Explanation:

a) Data and Calculations:

Locations  Cost of land     Cost of       Cost of mining      Time taken to

                                        equipment   and reclamation    mine the area

1                 $100,000      $10,000         $5,000 per day        30 days

2                 $35,000       $6,000          $4,500 per day        45 days

3                 $30,000       $7,500          $3,500 per day       120 days

4                 $40,500       $8,000          $7,000 per day        65 days

Locations  Cost of land     Cost of       Cost of mining                      Total

                                        equipment   and reclamation                    Costs

1                 $100,000      $10,000         $150,000 ($5,000 * 30)  $260,000

2                 $35,000       $6,000         $202,500 ($4,500 * 45)   $243,500

3                 $30,000       $7,500         $420,000 ($3,500 * 120)  $457,500

4                 $40,500       $8,000         $445,000 ($7,000 * 65)   $503,500

7 0
3 years ago
Explain the differences between an ideal gas and a real gas.
Annette [7]

Answer:

Ideal Gas

The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.

Real Gas

The molecules of real gas occupy space though they are small particles and also have volume.

Explanation:

I think i did this right

3 0
3 years ago
Phosphorus tribromide decomposes to form phosphorus and bromine, like this: 4pbr3 (g) → p4 (g) + 6br2 (g) write the pressure equ
FrozenT [24]
Chemical reaction: 4PBr₃(g) → P₄(g) + 6Br₂<span>(g).
</span>Pressure equilibrium constant (Kp) express the relationship between product pressures and reactant pressures. The partial pressures of gases are used to calculate pressure equilibrium constant.
Kp = (p(P₄)  · p(Br₂)⁶) ÷ p(PBr₃)⁴. 
p(P₄) - partial pressure of phosphorus.
p(Br₂) - partial pressure of bromine.
3 0
4 years ago
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