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3 years ago

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Based on the results you observed for the iodine test and Benedict’s test, is it better to detect enzyme activity by measuring t

he reduction in the amount of substrate (iodine test) or by measuring the increase in the amount of product (Benedict’s test)? EXPLAIN your answer!

1 answer:

dlinn [17]3 years ago

3 0

Answer:

Is better use the Benedict's test by the increase in the amount of the products if the enzyme is a reductase

Explanation:

The Benedict's test works by the reaction of the reducing sugars with the ion cupric of the reactive. If the enzyme is a reductase (degrades polysaccharides into bi o monosaccharides), it should cut the polysaccharide bond and the products would react with the Benedict's cupric ion

I hope you undestand me

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Which of the following is true of the gas constant, R?

NikAS [45]

Answer: I think it’s A sorry if I’m wrong

Explanation:

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3 years ago

Calculate the energy released in each of the following fusion reactions. Give your answers in MeV. (a) 2H + 3H → 4He + n 17.54 C

Vikentia [17]

Answer:

a) E = 17.55 MeV

b) E = 18.99 MeV

c) E = 3.29 MeV

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) E = 4.075 MeV

Explanation:

Energy Released, $E = \triangle M * 931.5$

$\triangle M = \sum M_{product} - \sum M_{reactant}$

Mass of 1H, $M_{H} = 1.007823$

Mass of 2H, $M_{2H} = 2.0141u$

Mass of 3H, $M_{3H} = 3.016 u$

Mass of Helium, $M_{4He} = 4.002602u$

Mass of Beryllium, $M_{7Be} = 7.01693 u$

Mass of neutron, $M_{n} = 1.008664 u$

a) $2H + 3H \rightarrow 4He + n$

$\triangle M = (4M_{He} + M_{n} ) - (2M_{H} + 3 M_{H} )\\\triangle M = ( 4.0026 + 1.008664) - (2.0141 + 3.016 )\\\triangle M = -0.01884u$

Energy released,

$E = -0.01884 * 931.5\\E = -17.55 Mev$

Energy released = 17.55 MeV

b) $4He + 4He \rightarrow 7Be + n$

$\triangle M = (M_{7Be} + M_{n} ) - (M_{4He} + M_{4He} )\\\triangle M = ( 7.01693 + 1.008664) - (4.002602 + 4.002602 )\\\triangle M = 0.02039 u$

Energy released,

$E = 0.02039 * 931.5\\E = 18.99 Mev$

c) $2H + 2H \rightarrow 3 He$ + n

$\triangle M = (M_{3He} + M_{n} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.008664) - (2.0141 + 2.0141 )\\\triangle M = -0.003536 u$

Energy released,

$E = -0.003536 * 931.5\\E = -3.29 Mev$

E = 3.29 MeV(Energy is released)

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) $2H + 2H \rightarrow 3H + 1H$

$\triangle M = (M_{3H} + M_{1H} ) - (M_{2H} + M_{2H} )\\\triangle M = ( 3.016 + 1.007825) - (2.0141 + 2.0141 )\\\triangle M = -0.00435 u$

$E = -0.00435 * 931.5\\E =-4.075 Mev$

E = 4.075 MeV ( Energy is released)

5 0

3 years ago

Whats the answer to this

Natasha2012 [34]

I think the answer you have selected is right, I'm not sure tho

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3 years ago

Which of the following equations is balanced?

stiv31 [10]

Answer:

A) 4P + 5O₂ → 2P₂O₅

Explanation:

A) 4P + 5O₂ → 2P₂O₅

This equation is balanced. There are four phosphorus and ten oxygen atoms are on both side of equation.

B) 5P + 4O₂ → 2P₄O₅

This equation is not balanced. There are five phosphorus and eight oxygen atoms on left, eight phosphorus ten oxygen on right side of equation.

C) 2P + O₂ → P₂O₅

This equation is not balanced. There are two phosphorus, two oxygen atoms on left and two phosphorus five oxygen on right side of equation.

D) 4P + 2O₂ → 2P₄O₅

This equation is not balanced. There are four phosphorus, four oxygen atoms on left and eight phosphorus ten oxygen on right side of equation.

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3 years ago

Which element has the least density at STP?

vladimir2022 [97]

Answer:

Helium, this noble gas has the least density at STP

Explanation:

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