What important role does surface tension play in everyday life?

Answers

Answer 1

Answer: * Some insects can walk on water.

* The spherical shape of the drops of water.

*water resistance to the penetration of foreign bodies.

hope this helps!

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The most easily polluted groundwater reservoirs are ______ aquifers?

Answers

Water-table aquifers.

The most easily polluted groundwater reservoirs are water-table aquifers.

Infiltration is the process by which rainwater becomes groundwater. In the water cycle, which involves primarily; evaporation, condensation and precipitation. Henceforth, after precipitation comes infiltration in the land areas leaving the rainwater aggregated in a single location which accumulates over time. Then this process is followed by runoff and subsurface flow by which water flows through other bodies of water, either in river, sea or ocean, even lake. Other rainwaters that has segregated and wasn’t able to transport itself stays in the process of infiltration, thus groundwater.

An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container. This is the equation: CO(g)+H2O(g)--><-- CO2(g) + H2(g). How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol?

Answers

Answer : The moles of $CO_2$ added will be 1.12 mole.

Solution : Given,

Moles of $CO$ and $H_2O$ at equilibrium = 0.200 mol

Moles of $CO_2$ and $H_2$ at equilibrium = 0.600 mol

First we have to calculate the concentration of $CO,H_2O,CO_2\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CO=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }H_2O=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }CO_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

$\text{Concentration of }H_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$K_c=((0.600)* (0.600))/((0.200)* (0.200))$

$K_c=9$

Now we have to calculate the moles of $CO_2$ added.

Let the moles of $CO_2$ added is 'x'.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initially 0.200 0.200 0.600 0.600

Added moles 0 0 x 0

Change +0.1 +0.1 -0.1 -0.1

Final 0.3 0.3 (0.5+x) 0.5

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$9=((0.5)* (0.5+x))/((0.3)* (0.3))$

$x=1.12mol$

Therefore, the moles of $CO_2$ added will be 1.12 mole.

$\boxed{0.{\text{3 mol}}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ are added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol.

Further Explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{P(g)}} + {\text{Q(g)}} \rightleftharpoons {\text{R(g)}} + {\text{S(g)}}$

Equilibrium constant is the constant that relates to the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for the general reaction is as follows:

${\text{K}}=\frac{{\left[ {\text{R}}\right]\left[ {\text{S}}\right]}}{{\left[{\text{P}} \right]\left[ {\text{Q}} \right]}}$

Here,

K is the equilibrium constant.

P and Q are the reactants.

R and S are the products.

The given reaction is as follows:

${\text{CO}}\left(g\right) + {{\text{H}}_2}{\text{O}}\left( g \right)\rightleftharpoons {\text{C}}{{\text{O}}_2}\left( g \right) + {{\text{H}}_2}\left( g \right)$

The expression for the equilibrium constant for the given reaction is as follows:

${\text{K = }}\frac{{\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]\left[ {{{\text{H}}_{\text{2}}}} \right]}}{{\left[ {{\text{CO}}} \right]\left[ {{{\text{H}}_2}{\text{O}}}\right]}}$ ......(1)

Here,

K is the equilibrium constant.

$\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]$ is the concentration of carbon dioxide.

$\left[{{{\text{H}}_{\text{2}}}} \right]$ is the concentration of hydrogen.

$\left[ {{\text{CO}}}\right]$ is the concentration of carbon monoxide.

$\left[ {{{\text{H}}_2}{\text{O}}}\right]$ is the concentration of water.

Substitute 0.600 mol/L for $\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]$ , 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}} \right]$ , 0.200 mol/L for $\left[ {{\text{CO}}}\right]$ and 0.200 mol/L for $\left[ {{{\text{H}}_2}{\text{O}}} \right]$ in equation (1).

$\begin{aligned}{\text{K }}&=\frac{{\left( {{\text{0}}{\text{.600 mol/L}}}\right)\left( {{\text{0}}{\text{.600 mol/L}}}\right)}}{{\left( {{\text{0}}{\text{.200 mol/L}}}\right)\left( {{\text{0}}{\text{.200 mol/L}}}\right)}}\n&= 9\n\end{aligned}$

The value of equilibrium constant comes out to be 9 and it remains the same for the given reaction.

Rearrange equation (1) to calculate .

$\left[{{\text{C}}{{\text{O}}_{\text{2}}}} \right]=\frac{{{\text{K}}\left( {\left[ {{\text{CO}}} \right]\left[ {{{\text{H}}_2}{\text{O}}} \right]} \right)}}{{\left[ {{{\text{H}}_{\text{2}}}} \right]}}$ ......(2)

Substitute 9 for K, 0.300 mol/L for $\left[{{\text{CO}}}\right]$ , 0.200 mol/L for $\left[{{{\text{H}}_2}{\text{O}}}\right]$ and 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}}\right]$ in equation (2).

$\begin{aligned}\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]&=\frac{{{\text{9}}\left( {{\text{0}}{\text{.300 mol/L}}} \right)\left( {{\text{0}}{\text{.200 mol/L}}} \right)}}{{{\text{0}}{\text{.600 mol/L}}}}\n&= 0.{\text{9 mol/L}}\n\end{aligned}$

Initially, 0.6 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ were present in a 1-L container. But now 0.9 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ are present in it. So the extra amount of ${\text{C}}{{\text{O}}_{\text{2}}}$ can be calculated as follows:

$\begin{aligned}{\text{Amount of C}}{{\text{O}}_{\text{2}}}{\text{ added}} &= 0.{\text{9 mol}} - 0.{\text{6 mol}}\n&= 0.{\text{3 mol}}\n\end{aligned}$

Therefore 0.3 moles of carbon dioxide are added in a 1-L container.

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: brainly.com/question/3467841

2. Complete equation for the dissociation of (aq): brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: CO, H2, CO2, H2O, 0.9 mol/L, 0.2 mol/L, 0.6 mol/L, 0.3 mol/L, K, carbon dioxide, water, hydrogen, carbon monoxide.

Elements that are characterized by the filling of p orbitals are classified as _____.a. Groups 3A through 8A
b. transition metals
c. inner transition metals
d. groups 1A and 2A

Answers

Answer : The correct option is, (a) Groups 3A through 8A

Explanation :

The general electronic configurations of :

Group 1A : $ns^1$

Group 2A : $ns^2$

Group 3A : $ns^2np^1$

Group 4A : $ns^2np^2$

Group 5A : $ns^2np^3$

Group 6A : $ns^2np^4$

Group 7A : $ns^2np^5$

Group 8A : $ns^2np^6$

Transition metal : $(n-1)d^((1-10))ns^((0-2))$

Inner transition metal (Lanthanoids) : $4f^((1-14))5d^((0-1))6s^2$

Inner transition metal (Actinoids) : $5f^((0-14))6d^((0-1))7s^2$

From the general electronic configurations, we conclude that the groups 3A through groups 8A elements that are characterized by the filling of p-orbitals.

Elements that are characterized by the filling of p orbitals are classified as $\boxed{{\text{a}}{\text{. Groups 3A through 8A}}}$ .

Further Explanation:

In order to make the study of numerous elements easier, these elements are arranged in a tabular form in increasing order of their atomic numbers. Such a tabular representation of elements is called a periodic table. Horizontal rows are called periods and vertical columns are called groups. A periodic table has 18 groups and 7 periods.

a. Groups 3A through 8A

The elements from group 3A to 8A has the general outermost electronic configuration of $n{s^2}n{p^(1 - 6)}$ . So the added electrons are to be filled in p orbitals.

b. Transition metals

These metals have the general valence configuration of $\left( {n - 1} \right){d^(1 - 10)}n{s^(0 - 2)}$ . This indicates that the added electrons enter either s or d orbitals.

c. Inner transition metals

These are classified as lanthanoids and actinoids. The general outermost configuration of lanthanoids is $4{f^(1 - 14)}5{d^(0 - 1)}6{s^2}$ while that of actinoids is $5{f^(0 - 14)}6{d^(0 - 1)}7{s^2}$ . In both cases, the added electron enters either d or f orbitals.

d. Groups 1A and 2A

The elements of group 1A have the general valence electronic configuration of $n{s^1}$ . It implies the last or valence electron enters in the s orbital. The group 2A elements have a general configuration of $n{s^2}$ . Here also the last electron enters the s orbital.

So elements from groups 3A to 8A are classified by the filling of p orbitals and therefore option a is correct.

Learn more:

Which ion was formed by providing the second ionization energy? brainly.com/question/1398705
Write a chemical equation representing the first ionization energy for lithium: brainly.com/question/5880605

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Periodic classification of elements

Keywords: periodic table, configuration, ns1, ns2, d, p, f, 3A, 8A, transition metals, inner transition metals, lanthanoids, actinoids, orbitals, 1A, 2A.

What biome could be characterized by large evergreen trees, acidic soil, lichens, and heavy winter snowfall?

Answers

The biome that could be characterized by large evergreen trees, acidic soil, lichens, and heavy winter snowfall is TAIGA. The Taiga experiences extremely cold or hot weather conditions. This is why the main seasons in the taiga are summer and winter. It is actually the largest biome in the world, and is located at the tip of Eurasia up to North America.

Answer:

Taiga

Explanation:

Calculate the percent ionization of nitrous acid in a solution that is 0.311 M in nitrous acid (HNO2) and 0.189 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.5 × 10^-4. Calculate the percent ionization of nitrous acid in a solution that is 0.311 M in nitrous acid (HNO2) and 0.189 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is
4.5 × 10^-4.

Answers

HNO2 =====> H+ + NO2-
Initial concentration = 0.311
C = -x,x,x
E = 0.311-x,x,x

KNO2 ====>K+ + NO2-
Initial concentration = 0.189
C= -0.189,0.189,0.189
E = 0,0.189,0.189

Choose the correct equation to represent the following statement (unbalanced chemical equation):Aluminum reacts with hydrochloric acid to yield aluminum chloride and hydrogen.

A. AlCl3 + H2 Al + HCl
B. Al + HCl AlCl3 + H2
C. Al HCl + AlCl3 + H2
D. Al + HCl AlCl3 + H2O

Answers

B. Al +HCI AICI3 +H2. I hope this helps

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