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Fofino [41]
3 years ago
10

A biochemist is attempting to replicate a chemical reaction that commonly takes place in cells. during the chemical reaction, st

arch is broken down into glucose. what will most likely happen if the amount of enzyme that catalyzes this reaction is increased?
a. the reaction will slow down.
b. the reaction will stop.
c. the reaction will speed up.
d. the reaction will take place at a higher temperature.
Chemistry
1 answer:
BartSMP [9]3 years ago
4 0
If the concentration of enzyme increases, then reaction should speed up (answer C).
enzyme speeds up a chemical reaction by lowering activation energy to provide an alternative pathway of reaction mechanism. assuming that enzyme which usually is the limiting factor, the increase of enzyme concentration should lead to more product formation i.e. maltose from the breakdown of starch.
the enzyme discussed here should be amylase.

higher temperature beyond optimum value enzyme found in our human body will usually least to denaturation of active site of enzyme. hence lead to loss of enzyme activity permanently.
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The atomic mass of hydrogen-1 is 1.008 and the atomic mass of helium-4 is 4.003. Find the difference in grams between 4 moles of
CaHeK987 [17]

4 moles of hydrogen  = 4 * 1.008 = 4.032 grams

1 mole of helium = 4.003 grams

Difference is  4.032 - 4.003

= 0.029 g

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Which of the following instruments has the highest pitch?<br> bass drum<br> flute<br> trombone
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Answer:

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Explanation:

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Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
Dovator [93]

Answer: 406 hours

Explanation:

Q=I\times t

where Q= quantity of electricity in coloumbs

I = current in amperes = 39.5 A

t= time in seconds = ?

The deposition of copper at cathode is represented by:

Cu^{2+}+2e^-\rightarrow Cu

96500\times 2=193000 Coloumb of electricity deposits 1 mole of copper

i.e. 63.5 g of copper is deposited by = 193000 Coloumb

Thus 19.0 kg or 19000 g of copper is deposited by = \frac{193000}{63.5}\times 19000=57748032 Coloumb

57748032=39.5\times t

t=1461975sec=406hours    (1hour=3600s)

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3 0
3 years ago
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

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Galina-37 [17]
Hi,

Two sodium atoms are needed to create an ionic bond with sulfur. 
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