All matter has __. A.) COLOR B.) MASS C.) VOLUME D.) WEIGHT

Which two chemical substances are needed to begin the photosynthesis factory process?

Rudiy27

<span>Carbon dioxide CO2 and water H2O. Through photosynthesis makes sugar C6H12O6.</span>

5 0

2 years ago

What do the properties of waves moving through aquatic environments tell us about the organisms living in those environments?

pshichka [43]

Assmksksmwmwosmeodmksnsks

7 0

2 years ago

The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151

vitfil [10]

Answer:- The natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for $^1^5^1_E_u$ and $^1^5^3_E_u$ as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of $^1^5^1_E_u$ as n then the abundance of $^1^5^3_E_u$ would be 1-n .

Let's plug in the values in the formula:

$151.964=150.919860(n)+152.921243(1-n)$

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

$151.964-152.921243=150.919860n-152.921243n$

$-0.957243=-2.001383n$

negative sign is on both sides so it is canceled:

$0.957243=2.001383n$

$n=\frac{0.957243}{2.001383}$

$n=0.478$

The abundance of $^1^5^1_E_u$ is 0.478 which is 47.8%.

The abundance of $^1^5^3_E_u$ is = $1-0.478$

= 0.522 which is 52.2%

Hence, the natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

3 0

2 years ago

professional and amateur athletes often tested for the use of illegal performance in Hensing drugs which process is used to sepa

Wittaler [7]

Answer:

Chromatography

Explanation:

7 0

3 years ago

Read 2 more answers

For an enzyme that displays Michaelis-Menten kinetics, what is the reaction velocity, v(as a percentage of Vmax), observed at:a)

fiasKO [112]

Answer:

a) 50% of the maximum velocity

b) 33.33% of the maximum velocity

c) 9.09% of the maximum velocity

d) 66.66% of the maximum velocity

e) 90.9% of the maximum velocity

Explanation:

The Michaelis-Menten kinetis is represented by

v = Vmax*S/(Km+S)

where

v= reaction rate

S= Substrate's concentration

Vmax= maximum rate of reaction

Km= constant

a) for S=Km

v = Vmax*Km/(2Km) = Vmax/2

v/Vmax = 1/2= 50% of the maximum velocity

b) for S=Km/2

v = Vmax*(Km/2)/(3/2Km) = Vmax/3

v/Vmax = 1/3= 33.33% of the maximum velocity

c) for S= 0.1*Km=Km/10

v = Vmax*(Km/10)/(11/10Km) = Vmax/11

v/Vmax = 1/11= 9.09% of the maximum velocity

d) for S=2*Km

v = Vmax*(2*Km)/(3*Km) = (2/3)* Vmax

v/Vmax = 2/3 = 66.66% of the maximum velocity

d) for S=10*Km

v = Vmax*(10*Km)/(11*Km) = (10/11)* Vmax

v/Vmax = 10/11 = 90.9 % of the maximum velocity

7 0

3 years ago