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2 years ago

Place the following in correct sequence from simplest to most complex: 1. molecules 2. atoms 3. tissues 4. cells 5. organs

Chemistry

1 answer:

Lera25 [3.4K]2 years ago

6 0

Answer:

Simplest to complex

Atoms, molecules, cells, tissues, organs

Explanation:

An atom is the smallest or basic unit of any substance or thing. therefore, it is simplest among given.

Two or more atoms joined together to form molecules. Therefore, molecules are complex than atoms.

Cell is the basic unit of life and a cell is made of many molecules and compounds. So, cells are more complex than molecules.

Several cells organised in a definite way to form tissue, so tissues are group of cells and perform a definite function inside a living thing. So, tissues are more complex than cell.

Organs consist of several cells and tissues and perform a specific functions. Therefore, organs are the most complex structure among given.

Therefore order from simplest to complex is as follows:

Atoms, molecules, cells, tissues, organs

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Type in the maximum number of electrons that can be present in each shell or subshell below.

diamong [38]

n = 5 shell=50

n = 2 shell=8

n = 2, l = 0 subshell=2

n = 2, l = 1 subshell=6

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2s subshell=2

5f subshell=14 sorry if I'm late I just did this! hope it helps..

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1 year ago

Read 2 more answers

What is the stoichiometric ratio between BaCl2 and NaCl

bixtya [17]

BaCl2+Na2SO4---->BaSO4+2NaCl There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent? "First convert grams into moles" 1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2 1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4 (7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2 "From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl" The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.

7 0

2 years ago

A sample of SO3 is introduced into an evacuated sealed container and heated to 600 K. The following equilibrium is established:

Andreas93 [3]

Answer: The value of $K_p$ is 0.050.

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_x=x\times P$

As we know the mole fraction of $O_2$ is 0.12

The partial pressure of $O_2=0.12\times 3.0atm=0.36atm$

The partial pressure of $SO_2=2\times 0.36atm=0.72atm$ Thus the partial pressure of $SO_3$ is = [3 - (0.36+0.720)] atm = 1.92 atm

$p_{SO3}$ = 1.92 atm

$2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)$

$K_p=\frac{p_{O_2}\times (p_{SO_}2)^2}{(p_{SO_3})^2}$

$K_p=\frac{0.36\times (0.72)^2}{(1.92)^2}$

$K_p=0.050$

The value of $K_p$ is 0.050.

7 0

2 years ago

When an alpha particle is emitted from an unstable nucleus, the atomic mass numberof the nucleus

barxatty [35]

Answer:

The correct option is D.

Explanation:

Radioactive substances usually emit different types of particles when they are decaying. Such particles include alpha particles, beta particles and gamma ray. When an alpha particle is emitted from an unstable radioactive nucleus such nucleus usually lost an atomic mass that correspond to that of helium atom. Note that an alpha particle is made up of two protons and two neutrons, which result in mass number of 4. Thus, a nucleus that emit an alpha particle will have its mass number (atomic mass) reduce by 4 and atomic number that is reduced by 2.

3 0

2 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

2 years ago