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3 years ago

What is the precision of 45.02

Chemistry

2 answers:

Tasya [4]3 years ago

8 0

Idk im very very sorry

gregori [183]3 years ago

6 0

Uhhhh hmmm im not sure

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3 years ago

What is the molar mass of CO2

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3 years ago

Which pair has identical electron configurations?

Thepotemich [5.8K]

Identical electron configurations : K⁺ and Cl⁻

<h3>Further explanation </h3>

In an atom, there are levels of energy in the shell and sub-shell

This energy level is expressed in the form of electron configurations.

Charging electrons in the sub-shell uses the following sequence:

S²⁻ and CI

S²⁻ : [Ne] 3s²3p⁶

Cl : [Ne] 3s²3p⁵

K⁺ and CI⁻

K⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶

Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶

S and Ar

S :[Ne] 3s²3p⁴

Ar : [Ne] 3s²3p⁶

Cl⁻ and K

Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶

K : 1s² 2s² 2p⁶ 3s² 3p⁶4s¹

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3 years ago

Do nonmetals form anions or cations?

MrRa [10]

Since medals form cations
nonmedals form anions

8 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

Tanzania [10]

Answer : The [H] is increasing at the rate of 0.36 mol/L.s

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

$\text{Rate of formation of }H=+\frac{d[H]}{dt}$

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0

3 years ago