Oxidation no of carbon in C3H7OH

marissa [1.9K]

Answer:

How does the equilibrium change with the removal of hydrogen (H2) gas from this equation? 2H2S ⇌ 2H2(g) + S2(g) A. ... Equilibrium shifts left to produce less reactant.

Explanation:

option A is the correct answer

Equilibrium shifts right to produce more product.

I hope it will help you.

3 0

2 years ago

Which of these pairs of elements have the same number of valence electrons?

Taya2010 [7]

The same number of valence electrons exists in the same family of elements in the periodic table. In finding the answer to this problem, we just have to find the pair that belongs to the same family in the periodic table. The same family means same column. Among these, Oxygen and lead (Pb) are in the same family. This pair is the answer.

7 0

3 years ago

Exposure of silver Chloride to sunlight for a long<br>duration turns grey due to

Tema [17]

Answer:

Exposure of silver chloride to sunlight for a long duration turns grey due to photolytic decomposition i.e decomposition in the presence of sunlight.

Explanation:

When silver chloride, AgCl is exposed to sunlight for a long time, it will undergo decomposition as the sunlight provides sufficient energy needed to decomposed the salt, AgCl to metallic silver and chlorine gas. This can be seen in the equation below:

2AgCl —> 2Ag + Cl2

5 0

3 years ago

For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

3 years ago

Which aqueous solution has the lowest freezing point c6h12o6, c2h5oh, ch3cooh, or nacl?

Eddi Din [679]

Depression of a freezing point of the solutions depends on the number of particles of the solute in the solution.
1 mol of C6H12O6 after dissolving in water still be 1 mol, because C6H12O6 does no dissociate in water.
1 mol of C2H5OH after dissolving in water still be 1 mol, because C2H5OH does no dissociate in water.
1 mol of NaCl after dissolving in water gives 2 mol of particles (ions), because NaCl is a strong electrolyte(as salt) and completely dissociates in water.
NaCl ----->Na⁺ + Cl⁻
1 mol of CH3COOH after dissolving in water gives more than 1 mol but less than 2 moles, because CH3COOH is a weak electrolyte (weak acid) and dissociates only partially.

So, most particles of the solute is going to be in the solution of NaCl,
so<span> the lowest freezing point has the aqueous solution of NaCl.</span>

3 0

3 years ago

Oxidation no of carbon in C3H7OH​

Oxidation no of carbon in C3H7OH