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3 years ago

Use the periodic to write the electron configuration for rubidium (Rb) in noble-gas notation. . Rb: [Kr]s

Chemistry

2 answers:

Mashcka [7]3 years ago

6 0

Answer:

the answer is 5

Explanation:

gladu [14]3 years ago

5 0

Answer:-

[Kr] 5s1

Explanation:-

Atomic number of rubidium = 37

Atomic number of Kr = 36

So only 1 electron needs to be assigned an orbital.

Kr is also the last element of the fourth period

So the shell number for the 1 electron will be 5 since it is after Kr

Since s subshell is filled first the electron will go into 5s.

Hence the electron configuration for rubidium (Rb) in noble-gas notation is

[Kr] 5s1

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The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pressures of H2O, H2, and O2

jenyasd209 [6]

Answer:

$1.742\times 10^{-5}$ is the value of the equilibrium constant at this temperature.

Explanation:

$2H_2O\rightleftharpoons 2H_2+O_2$

We are given:

Partial pressure of $H_2O=p^o_{H_2O}=0.0750 atm$

Partial pressure of $H_2=p^o_{H_2}=0.00700 atm$

Partial pressure of $O_2=p^o_{O_2}=0.00200 atm$

The expression of $K_p$ for the given chemical equation is:

$K_p=\frac{p^o_{H_2}^2\times p^o_{O_2}}{p^o_{H_2O}^2}$

Putting values in above equation, we get:

$K_p=\frac{(0.00700 atm)^2\times 0.00200 atm}{(0.0750 atm)^2}\\\\K_p=1.742\times 10^{-5}$

$1.742\times 10^{-5}$ is the value of the equilibrium constant at this temperature.

3 0

4 years ago

1. Define action energy.

WARRIOR [948]

Answer:

In physics, action is an attribute of the dynamics of a physical system from which the equations of motion of the system can be derived through the principle of stationary action. ... Action has dimensions of energy⋅time or momentum⋅length, and its SI unit is joule-second.

Explanation:

hope it helps

6 0

3 years ago

Read 2 more answers

2Mg + O2 → 2Mgo

Ray Of Light [21]

Answer:

it would be 4.5 mole left over

Explanation:

4 0

3 years ago

Calculate the enthalpy of the reaction

harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0

3 years ago

What is the velocity of a first-order reaction when the reactant concentration is 6 x 10-2M and the rate constant is 8 x 103sec-

tresset_1 [31]

Answer:

D) 4.8 × 10² M/s

Explanation:

Let's consider the following generic reaction.

A → B

The rate law is:

v = k . [A]ⁿ

where,

v: velocity

k: rate constant

[A]: concentration of the reactant A

n: reaction order for A

<em>What is the velocity of a first-order reaction when the reactant concentration is 6 × 10⁻²M and the rate constant is 8 × 10³s⁻¹?</em>

v = k . [A]¹ = 8 x 10³s⁻¹ . 6 x 10⁻²M = 4.8 × 10² M/s

5 0

4 years ago