1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
QveST [7]
3 years ago
14

Is the following chemical equation balanced? 2Na + 2H2O 2NaOH + H2

Chemistry
2 answers:
Mumz [18]3 years ago
8 0

yes because you have 2 H 2Na 2O on both sides


Harman [31]3 years ago
6 0

Answer : Yes, the given chemical reaction is balanced equation.

Explanation :

Balanced chemical reaction : It is defined as the number of atoms of individual elements present on reactant side must be equal to the product side.

The given chemical reaction is,

2Na+2H_2O\rightarrow 2NaOH+H_2

In this reaction, there are 2 atoms of sodium, 4 atoms of hydrogen and 2 atoms of oxygen are present on both side of the reaction. So,

This chemical reaction is a balanced reaction because in this reaction, the number of atoms of sodium, hydrogen and oxygen are balanced on the side of the reaction.

Hence, yes the given chemical reaction is balanced equation.

You might be interested in
Identify which result of a physical rather than a chemical change
Pani-rosa [81]

Explanation:

A physical change is change that alters the physical properties of matter especially its form and state.

In many cases, the change is easily reversible.

Examples are change of state such as boiling, melting, freezing, condensation, sublimation e.t.c

A chemical change is one in which a new kind of matter is formed. It is always accompanied by energy changes.

Examples are combustion, rusting , precipitation, milk souring.

  • Chemical changes are irreversible
  • Physical changes do not lead to the formation of new kinds of matter.
  • Most physical changes requires little energy.

You can use this knowledge to solve your problem

learn more:

Chemical change brainly.com/question/9388643

#learnwithBrainly

8 0
3 years ago
What is the purpose of sodium and chloride ions in the exoeruent?
Klio2033 [76]
What is the exoeruent. Searched it up on google and only came up with two search results. None related to chemistry
5 0
3 years ago
How many moles of bromine atoms are in 2.60×10^2 grams of bromine?
AlladinOne [14]
Your answer is 3.25 moles of Bromine

4 0
3 years ago
The Valence Electrons of an Atom of Which Element would feel a Greater Effective Nuclear Charge than the Valence?
kumpel [21]
The effective nuclear charge is an innate property of a specific element. It is the pull of force that an electron feels from the nucleus. It is related to the valence electron by the equation: Z* = Z-S, where Z* is the effective nuclear charge, Z is the atomic number and S is the shielding constant.

For the following elements in the choices, these are their values of Z*:

Aluminum - +12.591
Beryllium - +1.912
Hydrogen - +1
Carbon - +4

The effective nuclear charge of Boron is +3. Thus, the answers are  Aluminum and Carbon.
5 0
3 years ago
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
2 years ago
Other questions:
  • How many moles of PBr3 contain 3.68 x 10^25 bromine atoms?
    11·2 answers
  • A nurse is caring for a client who is receiving chemotherapy and has a platelet count of 30,000/mm3. Which statement by the clie
    11·1 answer
  • Photosynthesis uses 660-nm light to convert co2 and h2o into glucose and o2. calculate the frequency of this light.
    15·1 answer
  • A chemist determined by measurements that moles of calcium sulfide participate in a chemical reaction. Calculate the mass of cal
    12·1 answer
  • Please help!!
    7·1 answer
  • Can Individual atoms be moved and rearranged ?
    14·1 answer
  • Competition is most likely to occur between which two organism
    15·1 answer
  • ipt A wall that was once white is painted black. Which of the following is definitely true of the painted wall? A. Its chemical
    7·2 answers
  • What further observation led mendeleev to create the periodic table
    7·1 answer
  • The total energy radiated by a blackbody depends on.
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!