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jarptica [38.1K]
4 years ago
8

At temperatures above roughly 800 °c, calcium carbonate will decompose. write and balance the equation for this reaction. phase

symbols are optional.
Physics
2 answers:
Lady_Fox [76]4 years ago
8 0
CaCO3 ---->  CaO +  CO2 

hope it helps!

VladimirAG [237]4 years ago
7 0
<h3><u>Answer;</u></h3>

Decomposition of calcium carbonate

CaCO3(s) → CaO(s) + CO2(g)

<h3><u>Explanation;</u></h3>
  • <em><u>Decomposition reactions are types of reaction that occurs when a reactant is broken down into two or more products. </u></em>
  • Decomposition may either be thermal where heat is used to break a reactant or may be catalytic where a catalyst is used to break down a reactant to yield two or more substance, for example decomposition of hydrogen peroxide.
  • <em><u>All metal carbonates decompose to produce a metal oxide and carbon dioxide gas.</u></em>
  • <em><u>Calcium carbonate undergoes thermal decomposition to form calcium oxide and carbon dioxide. </u></em>
  • The equation will be;

CaCO3(s) → CaO(s) + CO2(g)

The equation is balanced as the number of atoms of each element are equal on both side.

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What is an electric field? How<br> do you calculate the electric<br> field due to a point charge?
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In a rectangular coordinate system, a positive point charge q = 6.50 nC is placed at the point x=0.190 m ,y=0, and an identical
son4ous [18]

Answer:

A) x and y components of the electric field  (Ep) at the origin.

Epx = -1620.5 N/C

Epy = -1620.5 N/C

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

   Ep= 2291.7 N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence  

1nC= 10⁻⁹C

Data

K= 9x10⁹N*m²/C²

q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C

d₁=d₂=0.190m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁ : Electric Field at point P  (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.

Ep₂: Electric Field at point  P (x=0, y=0) due to charge q₂. As the charge q₂  is positive (q₂+) ,the field leaves the charge

Ep: Total field at point P due to charges q₁ and q₂.

Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂

Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C

Look at the attached graphic :

Epx = Ep₁= -1620.5 N/C

Epy = Ep₂= -1620.5 N/C

A) x and y components of the electric field  (Ep) at the origin.

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

E_{p} =\sqrt{1620.5^{2}+1620.5^{2} } = 2291.7 \frac{N}{C}

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Using
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