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alex41 [277]
3 years ago
6

Enzymes are affected by both ph and

Chemistry
2 answers:
Musya8 [376]3 years ago
5 0
A. temperature
Enzymes can be effected by many factors but they usually vary down to these two
madreJ [45]3 years ago
3 0
Temperature where different enzymes work at different temperature
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NO LINKS PLEASE! Why does dye dropped into a cup of water eventually spread throughout the water?
Nezavi [6.7K]

Answer:

C

Explanation:

okay, you need to look at the structures of the particles of matter in the solid, liquid and gas.

  • particles in a solid are in fixed positions, where they can only vibrate in those positions ( take a look at ice, or rather, a brick)
  • liquids have very small or rather, no spaces between them, but they can slide or rub against each other, like people in a <em>really tight</em> crowd I guess
  • gas particles have very large spaces between them and they move randomly. these exibit what's called brownian motion.
  • since water particles (and all other liquid particles) have negligible spacings and limited movement, that allows the dye particles to move from a region of high concentration to that of a low concentration. the aim for this is for the mixture/solution to reach an equilibrium, that  is the mixture must get to a point where all regions have the same concentration of the dye.

you can refer to your coursebooks :)

correct where wrong please:)

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2 years ago
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Feliz [49]
Hi! How is your day?
3 0
2 years ago
Read 2 more answers
A gas has an initial volume of 455 mL at 105ºC and a final volume of 235 mL. What is its final temperature in Celsius degrees?
Oksana_A [137]
Hello!

To solve this problem we're going to use the Charles' Law. This Law describes the relationship between Volume and Temperature in an ideal gas. Applying this law we have the following equation:

\frac{V1}{T1} = \frac{V2}{T2} \\  \\ T2= \frac{V2*T1}{V1}=  \frac{235 mL * 105  ^{\circ}C }{455 mL}=54,23 ^{\circ}C

So, the final temperature is 54,23 °C

Have a nice day!
5 0
3 years ago
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A star is estimated to have a mass of 2.0 x 10 ^36kg. Assuming it to be a sphere of average radius of 7.0 x 10 ^5 km. Calculate
Montano1993 [528]

Answer:

<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>

<em>b) 8.69 x 10^7 lb/ft^3</em>

<em></em>

Explanation:

mass of the star m =  2.0 x 10^36 kg

radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m

The density of substance ρ = mass/volume

The volume of the star = volume of a sphere = \frac{4}{3}\pi  r^{3}

==> V = \frac{4}{3}*3.142*(7.0*10^8)^{3} = 1.437 x 10^27 m^3

density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3

in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>

in lb/ft^3 =  (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>

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3 years ago
How does evolution work
sergey [27]
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4 0
3 years ago
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