Ask question

Ask question

All categories

3 years ago

10

Put the level of organization in order from smallest to largest a. Organelles, cellsorgans, tissuesorgan systems b. Organs, tiss

ues, cells, organ systems, organelles Organelles, cells, tissues, organs, organ systems d. Organ systems, organsorganellestissues, cells

1 answer:

Lesechka [4]3 years ago

8 0

Answer: My guess is D i hope this is right

Explanation:

You might be interested in

What is true of neutrons?

Marrrta [24]

Neutron, neutral subatomic particle that is a constituent of every atomic nucleus except ordinary hydrogen. It has no electric charge and a rest mass equal to 1.67493 × 10−27 kg—marginally greater than that of the proton but nearly 1,839 times greater than that of the electron.

7 0

3 years ago

Which of the following actions will not increase the rate at which a solid dissolves in

aliya0001 [1]

Answer:

O lowering the temperature of the system

5 0

2 years ago

What is the correct answer?!????

ddd [48]

Answer:

B

Explanation:

Dalton worked with mainly about the chemistry of atoms.

how do atoms combine to form various molecules.

—rather than the details of the physical, internal structure of atoms, although he never denied the possibility of atoms' having a substructure.

4 0

3 years ago

The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

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.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$

The rate law expression will be:

$Rate=k[NO_2][O_3]$

Given:

Rate constant = $k=1.69\times 10^{-4}M^{-1}s^{-1}$

$[NO_2]$ = $1.77\times 10^{-8}M$

$[O_3]$ = $1.59\times 10^{-7}M$

$Rate=k[NO_2][O_3]$

$Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})$

$Rate=4.77\times 10^{-19}M/s$

The expression for rate of appearance of $NO_3$ :

$\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}$

As, $\text{Rate of reaction}=4.77\times 10^{-19}M/s$

So, $\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s$

Thus, the appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

7 0

3 years ago

madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0

3 years ago