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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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Scorpion4ik [409]

The answer would be D because a chemical property is Burning is a chemical property and a chemical property is any of material's properties that becomes evident during or after a chemical reaction that is any quality that can be established only by changing a substance's chemical identity.

i hope i helped you out and i hope you did great and i am sorry if i am too late

3 0
3 years ago
Read 2 more answers
How many moles of the product fecl3 do you make, when 8moles of fe react with excess cl2
tatuchka [14]

To figure out questions related to reacting moles/masses, the first step is always to write a complete balanced equation.

2Fe (s) + 3Cl2 (g) → 2FeCl3 (s)

Since Cl2 is the excess reactant, and Fe is the limiting reactant, we can simply find the number of moles of the product by comparing the mole ratio of the limiting reactant to the mole ratio of the product from the equation.

From the equation, mole ratio of Fe:FeCl3 = 2:2 = 1:1, the number of moles of product is exactly the same as the number of moles of the limiting reactant, which makes it 8 moles.

Note that if the mole ratio is not 1:1, you have to do some calculations to make sure the no. of moles is balanced at the end. Which means, if the mole ratio happened to be 1:2, the no. of moles of the product would be 8x2=16 instead.

So, your answer is 8 moles.

7 0
3 years ago
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FIRST ANSWER GETS CROWNED BRAINLIEST!!!!!!
Yakvenalex [24]

Answer:

75,900 g

Explanation:

m=d×v

(1.38g/cm^3) × (55,000)

7 0
4 years ago
Why can’t you see the stars very well at night like you could 50 years ago
Elenna [48]

Answer:Artificial light from cities has created a permanent "skyglow" at night, obscuring our view of the stars. Here's their map of artificial sky brightness in North America, represented as a ratio of "natural" nighttime sky brightness. In the black areas, the natural night sky is still (mostly) visible.

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6 0
2 years ago
The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The
dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e

Thus for mass number : 235 = 207+4X

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X = 7

Thus for atomic number : 92 = 82+2X-Y

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14-10 = Y

Y= 4

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+7_2^4\alpha+4_{-1}^0e

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

3 0
3 years ago
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