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Luba_88 [7]
3 years ago
5

Which molecules are pure elements?

Chemistry
2 answers:
Elenna [48]3 years ago
8 0
Well none since molecules are a group of two or more atoms electrically bonded with one another. However, there are gases that does not naturally bond due to their stability and can be found in nature as pure elements. But these are not considered as molecules.

(By the way, these gases are the noble gases that can be found on the last column of the periodic table) 
Vaselesa [24]3 years ago
4 0

Answer: the molecules formed by one kind of atom are pure elements


Explanation:


1) The definition of molecules is the covalent bonding of two or more atoms.


2) Under that definition you must have at least two atoms to be considered a molecule.


3) Pure elements are the 118 elements that you find in the periodic table: each contain only one kind of atom.


4) Then, molecules of pure elements, are two or more atoms of the same kind linked by covalent bonds.


5) With that these some examples of molecules that are pure elements:


i) O₂

ii) N₂

iii) F₂

iv) Cl₂

v) Br₂

vi) O₃


6) And here some examples of molecules that are not pure elements (there are many, many):


i) CH₄

ii) CH₃CH₂CH₃

ii) CH3CH₂CH₂CH₂CH₂CH₂CH₂CH₃


7) Note that ionic compounds are excluded of the definition of molecules, since they are not formed by covalent bonds but by ionic bonds. For example, NaCl is not a molecule.

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Consider the reaction: 2A(g)+B(g)→3C(g). When A is changing at a rate of -0.110M⋅s−1, How fast is C increasing?
mr_godi [17]
From the reaction above, the rate is given by the following formula:
r = -(1/2) dA / dt = - dB / dt = (1/3) dC/ dt
Note that A and B charge is negative due to they decrease with time
given dA / dt = -0.110 M/s
hence dB / dt = -0.110 / 2 = -0.055 M/s
dC / dt = (-3/2) (-0.110) = 0.165 M/s 
5 0
3 years ago
national defense, currency, post office, foreign affairs, & interstate commerce. b. charter local governments, education, pu
katen-ka-za [31]

Answer And Explanation:

Option C is correct.

Lend and borrow money, taxation, law enforcement, charter banks and transportation.

Some of the powers that were mentioned in the other options that weren't concurrent powers (that is, they belong to either the state government alone or the federal government alone) & disqualified them from being the answer include:

National defence (federal), Currency (federal), foreign affairs (federal), intrastate commerce (state) etc.

3 0
3 years ago
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hjlf

Answer:

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Explanation:

5 0
3 years ago
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A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
3 years ago
Open the Balancing Chemical Equations interactive and select Introduction mode. Then choose Separate Water. Adjust the coefficie
aleksklad [387]
<h2>H_2O  + H_2 + O_2</h2>

Explanation:

1. Water decomposition

  • Decomposition reactions are represented by-

       The general equation: AB → A + B.

  • Various methods used in the decomposition of water are -
  1. Electrolysis
  2. Photoelectrochemical water splitting
  3. Thermal decomposition of water
  4. Photocatalytic water splitting
  • Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.
  • The chemical equation will be -

        H_2O  + H_2 + O_2

Hence, balancing the equation we need to add a coefficient of 2 in front of H_2O on right-hand-side of the equation and  2 in front of H_2 on left-hand-side of the equation.

     ∴The balanced equation is -

       2 H_2O → 2 H_2 + O_2

2. Formation of ammonia

  • The formation of ammonia is by reacting nitrogen gas and hydrogen gas.

      N_2 + H → NH_3

Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.

   ∴The balanced chemical equation for the formation of ammonia gas is as  follows -

     N_2+3H→ 2NH_3.

  • When 6 moles of N_2 react with 6 moles ofH_2 4 moles of ammonia are produced.

5 0
3 years ago
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