All categories

1 year ago

Manganese (IV) perbromate please put into formula form

Chemistry

1 answer:

mamaluj [8]1 year ago

4 0

Answer

The formula form of Manganese (IV) perbromate is

$Mn(BrO_4)_4$

Explanation

The formula of Manganese is Mn

The formula for perbromate is BrO₄⁻

Oxidation number of Manganese (IV) = +4, That is Manganese (IV) is Mn⁺⁴

Therefore, multiply the charge of manganese by 1 and perchlorate by 4 t

You might be interested in

Which type of reaction involves the transfer of

baherus [9]

The type of reaction that involved the transfer of electrons is oxidation-reduction

7 0

2 years ago

Read 2 more answers

Write the following number in Standard (Numeric) Form. 5.8x10-5

DerKrebs [107]

Answer:

the answer of this question is 0.000058

4 0

3 years ago

What was wrong with Bohr’s model of the atom?

Montano1993 [528]

it works very well for atoms with only one electron, like H or He+, but not at all for multi-electron atoms.Bohr's model breaks down when applied to multi-electron atoms.

7 0

3 years ago

Just help me fill in the blank, thank you!

SVEN [57.7K]

Answer: metals, 88 , periodic table , luster , heat, electricity , thin, right, opposite , appearance , luster , malleable , ductile, opposite , shiny, moderately

Explanation:

Most of the elements in the periodic table are metals. Most of them are present on the left side of the periodic table. Elements to the far right of the periodic table are nonmetals. Elements which exhibit properties of both metals and nonmetals and are termed metalloids or semimetals.

Metals are defined as the elements which loose electrons to attain stable electronic configuration. They are hard ,malleable ( betaen to sheets) ,ductile ( drawn into wires) ,sonorous ( produce sound ) etc.

Non-metals are defined as the elements which gain electrons to attain stable electronic configuration. They are brittle and soft and are the poor conductors of heat and electricity.

Metalloids are defined as the elements which show properties of both metals and non-metals. Metalloids are typically semi-conductors, which means that they both insulate and conduct electricity.

3 0

3 years ago

A student prepared an equilibrium solution by mixing the following solutions:A.2.00 mL of 0.00250 M Fe(NO3)3B.5.00 mL of 0.00250

Crank

Explanation:

$Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}$

A. 2.00 mL of 0.00250 M $Fe(NO_3)_3$

Moles of ferric nitrate = n

Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)

Molarity of ferric nitrate = 0.00250 M

$n=0.00250 M\times 0.002 L=0.000005 mol$

B. 5.00 mL of 0.00250 M $KSCN$

Moles of KSCN = n'

Volume of KSCN = 5.00 ml = 0.005 L ( 1 mL=0.001 L)

Molarity of KSCN = 0.00250 M

$n'=0.00250 M\times 0.005 L=0.0000125 mol$

C. 3.00 mL of 0.050 M $HNO_3$

Moles of nitric acid = n''

Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)

Molarity of nitric acid = 0.050 M

$n=0.050 M\times 0.003 L=0.00015 mol$

After mixing A, B and C together and their respective initial concentration before reaction.

After mixing A, B and C together the volume of the solution becomes = V

V = 0.002 L=0.005 L+0.003 L= 0.010 L

Concentration of ferric nitrate :

$[Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M$

Concentration of ferric ions :

$[Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M$

Concentration of nitrate ions from ferric nitrate:

$[NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M$

Concentration of KSCN :

$[KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M$

Concentration of $SCN^-$ ions:

$[SCN^-]=1\times [KSCN]=0.00125 M$

Concentration of potassium ions:

$[K^+]=1\times [KSCN]=0.00125 M$

Concentration of nitric acid :

$[HNO_3]=\frac{0.00015 mol}{0.010 L}=0.015 M$

Concentration of hydrogen ion :

$[H^+]=1\times [HNO_3]=0.015 M$

Concentration of nitrate ions from nitric acid :

$[NO_3^{-}]=1\times [HNO_3]=0.0015 M$

Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M

$Fe^{3+}+SCN^-\rightleftharpoons Fe(NCS)^{2+}$

given concentration of $Fe(NCS)^{2+}$ at equilbrium = $3.6\times 10^{-5} M = 0.000036 M$

initially :

0.0005 M 0.00125 M 0

At equilibrium

(0.0005-0.000036) M (0.00125-0.000036) M 0.000036 M

0.000464 M 0.001214 M 0.000036 M

The expression of an equilibrium constant will be given as;

$K_c=\frac{[Fe(NCS)^{2+}]}{[Fe^{3+}][SCN^{-}]}$

$=\frac{0.000036 M}{0.000464 M\times 0.001214 M}=63.91$

The value for the equilibrium constant is 63.91.

6 0

3 years ago