What is the overall equation for the covalent bond in H2O . Is it H2 + O2=H2O OR. H2 + O = H2O

Compounds which are characterized as covalent are generally made up of elements found in which part of the periodic table

uranmaximum [27]

Elements that make up the compounds that are characterized as covalent are mostly in the right part of the periodic table.

Alkalyne metals and alkalyne earth metals, which are located in the left part of the table, and given that those elements form ions with relative ease they make up ioinic compounds.

Covalent compounds are made up mainly of a non metal element and a metal elements with low metallic properties.

The metallic character of the elements decrease as you go to the right of the periodic table, that is why you found those elements in the right part of the periodic talbe.

7 0

3 years ago

What is the solubility of m(oh)2 in a 0.202 m solution of m(no3)2? ksp 4.45*10^-12?

umka2103 [35]

M(NO₃)₂ fully separates into M²⁺ and NO₃ ²⁻ and M(OH)₂ partially separates as M²⁺ and 2OH⁻

M(NO₃)₂ → M²⁺ + 2NO₃²⁻

0.202 M 0.202 M

M(OH)₂(s) ↔ M²⁺ (aq) + 2OH⁻(aq)

I - -

C -X +X +2X

E X 2X

Ksp = [M²⁺ (aq)] [OH⁻(aq)]²

4.45 * 10∧-12 = (0.202 + X ) (2X)²

Since X is very small, (0.202 + X ) = 0.202

4.45 * 10-12 = 0.202 * 4X²

X = 2.347 × 10∧-6 M

Hence the solubility of M(OH)2 is 2.347 × 10∧-6 M

6 0

3 years ago

Consider the neutralization reaction 2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq) 2HNO3(aq)+Ba(OH)2(aq)⟶2H2O(l)+Ba(NO3)2(aq) A 0.1

Kipish [7]

Answer: The concentration of $HNO_3$ solution will be 0.094 M.

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HNO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$

We are given:

Conversion factor: 1L = 1000 mL

$n_1=1\\M_1=?M\\V_1=0.110L=110mL\\n_2=2\\M_2=0.100M\\V_2=51.9mL$

Putting values in above equation, we get:

$1\times M_1\times 110=2\times 0.100\times 51.9\\\\M_1=0.094M$

Hence, the concentration of $HNO_3$ solution will be 0.094 M.

7 0

3 years ago

2c2h2(l) 5o2(g) yeild 4co2(g) 2h2o(g) if the acetylene tank contains 37.0 mol of c2h2 and the oxygen tank contains 81.0 mol of o

Murrr4er [49]

2C2H2(l) +5O2(g)→ 4Co2(g) + 2H2O

The limiting reactant for reaction above is O2

explanation

The limiting reagent is determined using mole ratio of both reactant.

that is the mole ratio of C2H2:O2 which is 2:5 .

The mole ratio above implies that 37.0 mole of C2H2 needs 37.0 moles x5/2=92 moles of O2.

Since the available moles of O2 was 81.0 mole and 92 moles are required to completely react with C2H2 , O2 is the limiting reagent.

6 0

3 years ago

equilibrium lab analysis questions: use what you learned in this lab to answer the following questions in complete sentences. 1.

Margaret [11]

Answer:

See below

Explanation:

a) Adding HCl

HBTB is yellow in acid solution.

b) Adding NaOH

HBTB is blue in basic solution.

c) Explanation

HBTB is a weak acid in which the undissociated and ionized forms have different colours.

$\underbrace{\hbox{\text{HBTB }}}_{\hbox{\text{yellow}}} + \text{{ H$_{2}$O}}\rightleftharpoons \text{H$_{3}$O$^{+}$} +\underbrace{\hbox{\text{ BTB$^{-}$}}}_{\hbox{\text{blue}}}$

Around pH 7 , the indicator consists of roughly equal amounts of the yellow and blue forms, so it appears green

According to Le Châtelier's Principle, if you apply a stress to a system at equilibrium, it will respond in a way that will relieve the stress.

When you added HCl, you increased the concentration of H₃O⁺. The system responded in a way that would decrease the H₃O⁺. That is, the position of equilibrium shifted to the left and produced more of the yellow form.

When you added NaOH, the base removed some of the H₃O⁺. The system responded in a way that would increase the H₃O⁺. That is, the position of equilibrium shifted to the right and produced more of the blue form

7 0

3 years ago