When the distance from one object to another is changing the object is said to be in

Qual é o principal fator que indica que o organismo está adptado?

Hoochie [10]

Answer:

Explanation:

Any genetically based physiological, behavioral, or ecological trait that enables an organism to cope with, and to survive and reproduce in, its environment represents an adaptation.

Qualquer traço fisiológico, comportamental ou ecológico de base genética que permite a um organismo enfrentar, sobreviver e se reproduzir em seu ambiente representa uma adaptação.

5 0

2 years ago

If the mass of a material is 40 grams and the volume of the material is 9 cm^3, what would the density of the material be?

solniwko [45]

Mass = m = 40 grams
Volume = V = 9 cm³
Density = d = ?

Density is defined as the ratio of mass and volume.

So,

d = m/V

Using the values, we get

d = 40/9 = 4.44 g/cm³

This means the density of material would be 4.44 g/cm³

3 0

2 years ago

Calculate the average atomic mass of Mg if the naturally occurring isotopes are Mg - 24, Mg - 25 and Mg - 26. Their masses and a

Elenna [48]

Answer:

24.309

Explanation:

mass×percentage+.. = average amu

23.985×78.7/100+24.986×10.13/100+25.983×11.17/100

to three decimal places

=24.309

5 0

1 year ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$