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3 years ago

Planets close to a star will have __ orbital velocities than planets farther from a star.

Chemistry

1 answer:

Naddika [18.5K]3 years ago

5 0

Answer:

The closer a planet is to the Sun, the stronger the Sun's gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun's gravitational pull, and the slower it moves in its orbit.

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Location E is located at the Tropic of Cancer, while location F is at the equator. Which location is likely to be warmer at the

Stells [14]

The answer is "Location E, because Earth's axis is tilted towards the sun" :D

6 0

2 years ago

Th e enzyme-catalyzed conversion of a substrate at 25°C has a Michaelis constant of 0.045 mol dm−3. Th e rate of the reaction is

Elis [28]

Answer:

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

Explanation:

Michaelis–Menten 's equation:

$v=V_{max}\times \frac{[S]}{K_m+[S]}=k_{cat}[E_o]\times \frac{[S]}{K_m+[S]}$

$V_{max}=k_{cat}[E_o]$

v = rate of formation of products =

[S] = Concatenation of substrate

$[K_m]$ = Michaelis constant

$V_{max}$ = Maximum rate achieved

$k_{cat}$ = Catalytic rate of the system

$[E_o]$ = Initial concentration of enzyme

We have :

$K_m=0.045 mol/dm^3$

$v=1.15 mmol/dm^3 s=1.15\times 10^{-3} mol/dm^3 s$

$[S]=0.110 mol/dm^3$

$v=V_{max}\times \frac{[S]}{K_m+[S]}$

$1.15\times 10^{-3} mol/dm^3 s=V_{max}\times \frac{0.110 mol/dm^3}{[(0.045 mol/dm^3)+(0.110 mol/dm^3)]}$

$V_{max}=\frac{1.15\times 10^{-3} mol/dm^3 s\times [(0.045 mol/dm^3)+(0.110 mol/dm^3)]}{0.110 mol/dm^3}=1.620\times 10^{-3} mol/dm^3 s$

$1.620\times 10^{-3} mol/dm^3 s$ is the maximum velocity of this reaction.

8 0

3 years ago

Three isotopes of argon occur in nature. Calculate the average atomic mass of argon given the following relative atomic masses a

Volgvan

Answer:

whats the question

Explanation:

6 0

3 years ago

Read 2 more answers

Question 15 need this answered asap (chemistry)

Tcecarenko [31]

Answer:

80.7 L

Explanation:

PV = nRT

P = 1520 mmHg = 2 atm

n = 5 mol

R = 0.08206 (L * atm)/(mol * K)

T = 393.15 K

2 (V) = 5 (0.08206) (393.15)

V ≈ 80.7 L

8 0

3 years ago

One and a half gallons of paint are used on refrigeration units which are then baked dry in a 275F oven for 2 hours. If the pain

brilliants [131]

Answer:

Hence the evaporation rate in pints per minute is 0.001583.

Explanation:

Now,

1.5 gallons of paint are used.

It has 5% solid, so total 95% of 1.5 gallon can evaporate(only the solvent part will evaporate).

95% of 1.5 gallon=(95/100)*1.5 = 1.425 gallons.

1gallon = 8 pints.

So 1.425 gallons=8*1.425 pints

=11.4 pints.

Evaporation requires 2 hours. That is 3600*2 seconds = 7200 seconds.

So evaporation rate= 11.4 pints/7200 seconds.

=0.001583 pints per second

4 0

3 years ago