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2 years ago

How many valence electrons does Rb (rubidium) have?

Chemistry

1 answer:

Murljashka [212]2 years ago

3 0

Answer:

The total number of electrons in a valence shell is called valence electrons. The electron configuration of rubidium shows that the last shell of rubidium has an electron. Therefore, the valence electrons of rubidium(Rb) are one.

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What kind of weather does a stationary front bring?

Margaret [11]

Same as a warm front drizzle rain or light snow, I think

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3 years ago

Read 2 more answers

The two common chlorides of phosphorus, PCl3, and PCl5, both important for the production of the other phosphorus compounds, coe

Irina18 [472]

Answer:

For a: The value of $K_c$ for the given reaction is 271.6

For b: The value of $K_p$ for the reaction is 6.32

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $PCl_5$ :

Given mass of $PCl_5$ = 0.105 g

Molar mass of $PCl_5$ = 208.24 g/mol

Putting values in equation 1, we get:

$\text{Moles of }PCl_5=\frac{0.105g}{208.24g/mol}=5.04\times 10^{-4}mol$

For $PCl_3$ :

Given mass of $PCl_3$ = 0.220 g

Molar mass of $PCl_5$ = 137.33 g/mol

Putting values in equation 1, we get:

$\text{Moles of }PCl_3=\frac{0.220g}{137.33g/mol}=1.60\times 10^{-3}mol$

For $Cl_2$ :

Given mass of $Cl_2$ = 2.12 g

Molar mass of $Cl_2$ = 71.0 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Cl_2=\frac{2.12g}{71.0g/mol}=0.029mol$

Volume of the flask = 25.0 L

For the given chemical equation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

For a:

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The expression of $K_c$ for above reaction follows:

$K_c=\frac{PCl_5}{PCl_3\times Cl_2}$

We are given:

$[PCl_5]=\frac{5.04\times 10^{-4}mol}{25L}$

$[PCl_3]=\frac{1.60\times 10^{-3}mol}{25L}$

$[Cl_2]=\frac{0.029mol}{25L}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{5.04\times 10^{-4}}{25})}{(\frac{1.60\times 10^{-3}}{25})\times (\frac{0.029}{25})}\\\\K_c=271.6$

Hence, the value of $K_c$ for the given reaction is 271.6

For b:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = 271.6

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $250^oC=250+273=523K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=1-2=-1$

Putting values in above equation, we get:

$K_p=271.6\times (0.0821\times 523)^{-1}\\\\K_p=6.32$

Hence, the value of $K_p$ for the reaction is 6.32

7 0

4 years ago

The name of the following molecular compound: P203

topjm [15]

B Phosphate trioxide

5 0

3 years ago

Write a chemical equation for the dissolution of the AgCl precipitate upon the addition of NH,(aq).

Kobotan [32]

Explanation:

White precipitate of silver chloride get dissolves in excess ammonia to formation of complex between silver ions, chloride ions and ammonia molecules.

The chemical reaction is given as:

$AgCl(s)+2NH_3(aq)\rightarrow Ag[(NH_3)_2]^+Cl^-(aq)$

When 1 mole of silver chloride is added to 2 mole of an aqueous ammonia it form coordination complex of diaaminesilver(I) chloride.

7 0

3 years ago

What is the molarity of a solution containing 0.5 mol of zinc sulfate (ZnSO4) per liter?

QveST [7]

molarity = G/L/RFM

RFM = 120+32+48= 200

0.5/200 = 0.0025

7 0

4 years ago