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AVprozaik [17]
3 years ago
10

Does the reaction of a main-group metal oxide in water produce an acidic solution or a basic solution? Write a balanced equation

for the reaction of a Group 2A(2) oxide with water.
Chemistry
1 answer:
Fynjy0 [20]3 years ago
4 0

<u>Answer:</u> The main group metal produce a basic solution in water and the reaction is MO+H_2O\rightarrow M(OH)_2

<u>Explanation:</u>

Main group elements are the elements that are present in s-block and p-block.

The metals that are the main group elements are located in Group IA, Group II A and Group III A.

Oxides are formed when a metal or a non-metal reacts with oxygen molecule. There are two types of oxides which are formed: Acidic oxides and basic oxides.

  • Acidic oxides are formed by the non-metals.
  • Basic oxides are formed by the metals.

When a metal oxide is reacted with water, it leads to the formation of a base.

The general formula of the oxide formed by Group II-A metals is 'MO'

The chemical equation for the reaction of metal oxide of Group II-A and water follows:

MO+H_2O\rightarrow M(OH)_2

Hence, the main group metal produce a basic solution in water and the reaction is MO+H_2O\rightarrow M(OH)_2

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Answer:

=> 1366.120 g/mL.

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m)     = 25 Kg

Volume (v) = 18.3 mL.

From our question, we are to determine the density (rho) of the rock.

The formula:

p = \frac{m}{v}

First let's convert 25 Kg to g;

1 Kg    = 1000 g

25 Kg = ?

= \frac{25 × 1000}{1}

= 25000 g

Substitute the values into the formula:

p =  \frac{25000 g}{18.3 ml}

= 1366.120 g/mL.

Therefore, the density (rho) of the rock is  1366.120 g/mL.

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How do you calculate the density of a gas, based on its temperature and pressure?
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A substance undergoes a change. Which of the following indicates that the change was a chemical change?
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2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s
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Answer:

In order to prepare 10 mL, 5 μM; <em> 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.</em>

In order to prepare 10 mL, 10 μM; <em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM; <em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM; <em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

                       = 2 mL

<em>2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

<em />

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

<em>4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

<em>6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

<em>8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.</em>

6 0
3 years ago
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