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3 years ago

Compared to an endocrine response a nerve response differs in that the nerve response generally.

Chemistry

1 answer:

Nadusha1986 [10]3 years ago

7 0

the nerve response is generally faster

Explanation:

nerves travel throughout neurotransmitters which is a faster response, like a reflex, than the hormones in the endocrine system, like the growth hormone.

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Which statement is true regarding a catalyst?

Alla [95]

Answer:

d). A catalyst lowers the activation energy for a chemical reaction.

Explanation:

From the given choices, it is true that a catalyst lowers the activation energy for a chemical reaction.

The activation energy is the energy barrier that must be overcome before a chemical reaction can occur.

Some reactions have very high activation energy and would not occur without the introduction of a catalyst.

The catalyst brings the reactants into contact by removing the energy deficiency in the system.

4 0

3 years ago

Which statements are true of fossil fuels?

Arisa [49]

Letters A and B are correct because burning fossil fuels can harm the atmosphere through the greenhouse effect and contribute to global warming, and can sometimes spill into the water and harm the environment.

7 0

3 years ago

Rank the elements in each of the following sets in order of increasing IE₁:

zavuch27 [327]

Explanation:

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period.

But when we move from top to bottom in a group then there occurs an increase in size of the atoms. Hence, ionization energy decreases along a group.

(a) As Sb, Sn and I are all period 5 elements. Hence, these elements are arranged in order of increasing $IE_{1}$ as follows.

Sn < Sb < I

(b) As Sr, Ca, and Ba are all elements of group 2a. Hence, these elements are arranged in order of increasing $IE_{1}$ as follows.

Ba < Sr < Ca

3 0

3 years ago

How many L of 3.0 M H2SO4 solution can be prepared by using 100.0 mL OF 18 M H2SO4?

max2010maxim [7]

Answer: A volume of 600 mL of 3.0 M $H_{2}SO_{4}$ solution can be prepared by using 100.0 mL OF 18 M $H_{2}SO_{4}$ .

Explanation:

Given: $V_{1}$ = ?, $M_{1} = 3.0 M\\$

$V_{2} = 100.0 mL$ , $M_{2} = 18 M$

Formula used to calculate the volume is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL$

Thus, we can conclude that a volume of 600 mL of 3.0 M $H_{2}SO_{4}$ solution can be prepared by using 100.0 mL OF 18 M $H_{2}SO_{4}$ .

6 0

3 years ago

Reaction rate is expressed in terms of changes in the concentration of reactants and products. Write a balanced equation for the

KengaRu [80]

Answer : The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

$Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}$

The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

5 0

3 years ago