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dimulka [17.4K]
3 years ago
11

What is the formula of hexaaquamanganese(ii) sulfate? express your answer as a chemical formula?

Chemistry
1 answer:
Digiron [165]3 years ago
3 0

Answer : The formula of hexaaquamanganese(II)sulfate is [Mn(H_2O)_6]SO_4.

Explanation :

Rules for writing formulas of coordination complexes :

  • Central metal is written first and then the ligands are written along with their prefixes.
  • The anion ligands are written before the neutral ligands.
  • If there are more than one anion ligands or neutral ligands then they are written in alphabetical order.

The given name of complex is, hexaaquamanganese(ii) sulfate.

In this, the central metal is manganese which is written as Mn and the ligand is water (which is a neutral ligand) is written as H_2O and an anion is sulfate which is written as SO_4.

Water is a neutral ligand and sulphate has a charge of -2. As the overall complex is neutral the metal bears a +2 charge which is represented in Roman numeral after the metal.

Therefore, the formula of hexaaquamanganese(II)sulfate is written as [Mn(H_2O)_6]SO_4.

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Formed when chemicals in the air get into rain and up the acidity levels
snow_lady [41]

Acid Rain is formed when chemicals in the air get into rain and up the acidity levels!!!!!!

Hope this helps guys!

4 0
2 years ago
How much (mL) of a 2.0 M sodium hydroxide solution would it take to neutralize 50 mL of a 6.0 M solution of hydrochloric acid?
Tom [10]

Answer:

150ml

Explanation:

For this question,

NaOH completely dissociates. It is a strong base

HCl also completely dissociates. It is a strong acid

So we have this equation

m1v1 = m2v2 ----> equation 1

M2 = 2m

V1= ??

M2 = 6m

V2 = 50m

When we input these into equation 1, we have:

2m x v1 = 6m x 50ml

V1 = 6m x 50ml/2

V1 = 300/2

V1 = 150ml

Therefore NaOH that is required to neutralize the solution of hydrochloric acid is 150ml.

Thank you

8 0
2 years ago
In the following reaction, what element is gaining mass?
vagabundo [1.1K]

Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).

8 0
2 years ago
A graduated cylinder was filled with 25.0 mL of water. A solid object was immersed in the cylinder, raising the water level to 3
yawa3891 [41]

Answer:

Density of the object is 8759.494 grams/L

Explanation:

As we know density of an object is mass of the object divided by its volume

Given

Volume of the object is equal to the change in volume of water with in the cylinder when the object was immersed in water.

Hence, volume of object is equal to

32.9 -25 = 7.9 mL

Mass of the object is 69.2 grams

Density

= \frac{69.2}{7.9 *10^{-3}}\\= 8759.494 grams/L

Density of the object is 8759.494 grams/L

7 0
3 years ago
Volume occupied 3.52x10^32 moluchles<br>of Mathane (CH4)<br>1) At STP​
Naddika [18.5K]

Answer:

volume = 13097674418.528dm³

Explanation:

n = (3.52)*10^32/(6.02)*10^23)

n = (584717607.97)

n = volume /molar volume

molar volume at stp = 22.4dm³

volume= 584717607.97 x 22.4

volume = 13097674418.528dm³

6 0
2 years ago
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