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miskamm [114]
2 years ago
9

Which of the following facts about the ocean water indicates that the ocean is a major resource of nutrients?

Chemistry
1 answer:
xz_007 [3.2K]2 years ago
7 0
I believe it’s B
Have a good day
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Temperature is _____ to the average kinetic energy of a gas.
d1i1m1o1n [39]

Answer: Option (b) is the correct answer.

Explanation:

As on increasing the temperature, the molecules gain more kinetic energy due to which they tend to collide and move rapidly from one place to another.  

Thus, we can conclude that when temperature is increased, the kinetic energy of the molecules increases.

This means that temperature is directly proportional to the average kinetic energy of a gas.

8 0
3 years ago
Read 2 more answers
What are deltaTb and deltaTf for an aqueous solution that is 1.5g nacl in 0.250kg h2o? Given Kb=0.51 C/m and kr=1.86 C/m
bulgar [2K]

Answer:

T_f for given question is 2.79 and T_b is 0.52

\Delta T_b = I \times K_b \times m {i- vant hoff’s constant ; Kb- constant ; m molarity }

M = no. of moles of the solute present in one kg of solution

Let the weight of amount of solute be “w” and its molecular mass be “M”

Let the mass of the solvent in the given question be “x”

\Delta T_b = I \times K_b \times (w/M)/ x

\Delta T_b = I \times K_b \times w/Mx

\Delta T_b = 1 \times 0.51 \times1.5/(0.250 \times 58.44) = 0.052

\Delta T_f = M \times K_f = 1.86 \times 1.5 = 2.79

4 0
3 years ago
In acetyl CoA formation, the carbon-containing compound from glycolysis is oxidized to produce acetyl CoA. From the following co
kondor19780726 [428]

Answer:

- Net Input: NAD⁺, coenzyme A, pyruvate

- Net Output: NADH, acetyl CoA, CO₂

- Not input or output: O₂, ADP, glucose and ATP

Explanation:

Hello,

In this case, it is important to recall that acetyl-CoA is produced either by oxidative decarboxylation of pyruvate derived from glycolysis, which is carried out into the mitochondrial matrix, by cause of the oxidation of high-order fatty acids, or by oxidative degradation of very specific amino acids. Acetyl-CoA then enters in the citric acid cycle where it is oxidized in the light of energy production.

In this manner, during such processes, there are some net inputs and outputs, therefore, they are sorted as show below, considering there some of them not classified neither as input nor output:

- Net Input: NAD⁺, coenzyme A, pyruvate

- Net Output: NADH, acetyl CoA, CO₂

- Not input or output: O₂, ADP, glucose and ATP

Best regards.

8 0
3 years ago
What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i
Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314  (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0
3 years ago
Consider the reaction NOBr(g) => NO(g) + 1/2 Br2(g). A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M-1
timama [110]

Answer:

Thus, the order of the reaction is 2.

The rate constant of the graph which is :- 2.00 M⁻¹s⁻¹

Explanation:

The kinetics of a reaction can be known graphically by plotting the concentration vs time experimental data on a sheet of graph.

The concentration vs time graph of zero order reactions is linear with negative slope.

The concentration vs time graph for a first order reactions is a exponential curve.  For first order kinetics the graph between the natural logarithm of the concentration vs time comes out to be a straight graph with negative slope.

The concentration vs time graph for a second order reaction is a hyberbolic curve. Also, for second order kinetics the graph between the reciprocal of the concentration vs time comes out to be a straight graph with positive slope.

Considering the question,

A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M⁻¹s⁻¹.

<u>Thus, the order of the reaction is 2.</u>

<u>Also, slope is the rate constant of the graph which is :- 2.00 M⁻¹s⁻¹</u>

5 0
3 years ago
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