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3 years ago

I will give brainliest!!

Chemistry

1 answer:

stich3 [128]3 years ago

5 0

Answer:

Explanation:

Biodiversity is the variety and variability of life on Earth. Biodiversity is typically a measure of variation at the genetic, species, and ecosystem level. Terrestrial biodiversity is usually greater near the equator, which is the result of the warm climate and high primary productivity. Biodiversity is not distributed evenly on Earth, and is richest in the tropics. These tropical forest ecosystems cover less than 10 percent of earth's surface, and contain about 90 percent of the world's species. Marine biodiversity is usually highest along coasts in the Western Pacific, where sea surface temperature is highest, and in the mid-latitudinal band in all oceans. There are latitudinal gradients in species diversity. Biodiversity generally tends to cluster in hotspots, and has been increasing through time, but will be likely to slow in the future.

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How can you form new neural pathways that will lead to transformed thinking and a transformed life?

aleksandr82 [10.1K]

When you understand how neural pathways are created in the brain, you ... But because I had the will to do it, I built a new pathway, and I rewired or reprogrammed my brain. ... can learn new behaviors and attitudes and can transform their lives. ... to the habit, and see what results you're creating in your life

6 0

3 years ago

The following mechanism has been suggested for the reaction between nitrogen monoxide and oxygen: NO(g) + NO(g) → N2O2(g) (fast)

Karo-lina-s [1.5K]

Answer:

b. Second order in NO and first order in O₂.

Explanation:

A. The mechanism

$\rm 2NO\xrightarrow[k_{-1}]{k_{1}}N_{2}O_{2} \, (fast)\\\rm N_{2}O_{2} + O_{2}\xrightarrow{k_{2}} 2NO_{2} \, (slow)$

B. The rate expressions

$-\dfrac{\text{d[NO]} }{\text{d}t} = k_{1}[\text{NO]}^{2} - k_{-1} [\text{N}_{2}\text{O}_{2}]^{2}\\\\\rm -\dfrac{\text{d[N$_{2}$O$_{2}$]}}{\text{d}t} = -\dfrac{\text{d[O$_{2}$]}}{\text{d}t} = k_{2}[ N_{2}O_{2}][O_{2}] - k_{1} [NO]^{2}\\\\\dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= k_{2}[ N_{2}O_{2}][O_{2}]$

The last expression is the rate law for the slow step. However, it contains the intermediate N₂O₂, so it can't be the final answer.

C. Assume the first step is an equilibrium

If the first step is an equilibrium, the rates of the forward and reverse reactions are equal. The equilibrium is only slightly perturbed by the slow leaking away of N₂O₂ to form product.

$\rm k_{1}[NO]^{2} = k_{-1} [N_{2}O_{2}]\\\\\rm [N_{2}O_{2}] = \dfrac{k_{1}}{k_{-1}}[NO]^{2}$

D. Substitute this concentration into the rate law

$\rm \dfrac{\text{d[NO$_{2}$]}}{\text{d}t}= \dfrac{k_{2}k_{1}}{k_{-1}}[NO]^{2} [O_{2}] = k[NO]^{2} [O_{2}]$

The reaction is second order in NO and first order in O₂.

8 0

3 years ago

For a ternary solution at constant T and P, the composition dependence of molar property M is given by: M = x1M1 + x2M2 + x3M3 +

AveGali [126]

Answer:

$M_{i} = M_{i} + C_{xjxk} (1-2x_{i})$ ...1

$M^{\alpha } = M_{i} + CX_{xjxk}$ ...2

Explanation:

The ternary constant is given by the following equation:

The symbol XiXi, where XX is an extensive property of a homogeneous mixture and the subscript ii identifies a constituent species of the mixture, denotes the partial molar quantity of species ii defined by

$M_{i} = [\frac{d(nM)}{dn_{i} }]_{P,t,n,j}$

This is the rate at which property X changes with the amount of species i added to the mixture as the temperature, the pressure, and the amounts of all other species are kept constant. A partial molar quantity is an intensive state function. Its value depends on the temperature, pressure, and composition of the mixture.

In a multi phase system (in this case, a ternary system), the components resolved give:

$M_{i} = M_{i} + C_{xjxk} (1-2x_{i})$

and $M^{\alpha } = M_{i} + CX_{xjxk}$

5 0

3 years ago

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) The Δ G ° ′ of the reaction is − 9.130 kJ ⋅ mol − 1 . Calculate the equil

Liula [17]

Answer:

K = 39.85

ΔG= -6.9 kJ/mol

Explanation:

Step 1: Data given

The ΔG°′ of the reaction is − 9.130 kJ/mol

Temperature = 25.0 °C = 298 K

Body temperature = 37.0 °C = 310K

the concentration of A is 1.9 M

the concentration of B is 0.80 M

Step 2: The reaction

A (aq) ⇌ B (aq)

Step 3:

ΔG° = -RT ln K

⇒with ΔG° = standard Gibbs free energy change = − 9.130 kJ/mol

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T = the temperature = 298 K

⇒with K = the equilibrium constant = TO BE DETERMINED

− 9130 J/mol = - 8.314 * 298 * ln K

ln K = 3.685

K = e^3.685

K = 39.85

Step 4: The reaction at body temperature

ΔG= ΔG° + RT ln [B]/[A]

⇒with ΔG° = Gibbs free energy change

⇒with ΔG° = standard Gibbs free energy change = − 9130 J/mol

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T = the temperature = 310 K

⇒with [A] = 1.9 M

⇒with [B] = 0.80 M

ΔG= -9130 J/mol + 8.314 J/mol*K * 310 K * ln (1.9/0.80)

ΔG= -9130 J/mol + 2229.4J/mol

ΔG=-6900.6 J/mol = -6.9 kJ/mol

6 0

3 years ago

A sample of gas has a volume of 1.9L and a temperature of 21 degrees celsius. Heat is applied to the sample, leading to an incre

Mashutka [201]

Answer:

1.94 L

Explanation:

21°C = 21 +273 = 294 K

27°C = 27 + 273 = 300 K

T1/V1 = T2/V2

294 K/1.9 L = 300 K/x L

x = (1.9*300)/294 ≈ 1.94 L

3 0

3 years ago