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jarptica [38.1K]

3 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.

2 answers:

Lady_Fox [76]3 years ago

8 0

CaCO3 ----> CaO + CO2

hope it helps!

VladimirAG [237]3 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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A 75kg man climbs the stairs to the fifth floor of a building. A total hieght of 16m. His potential energy has increased by

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Answer:

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Explanation:

Given

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3 years ago

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1. a=Δv/Δt=(v-vo)/t=(0-25)/5=-25/5=-5 m/s²

The "-" sign shows us that the car has a slow motion

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3 years ago

A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. (

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Answer:

A) $h = 69.58 m$

D) $v = 58.12 \frac{m}{s}$ (Speed magnitude)

α = 22.49° (Speed direction above the horizontal)

Explanation:

Conceptual analysis:

To solve this problem we consider the following concepts:

1) The projectile in its movement describes a curved line called a parabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane.

The initial velocity (Vo) is tangent to the trajectory at the initial point and can be broken down into two components, one vertical (Voy) and one horizontal(Vox):

$V_{ox} = V_{o}Cos\alpha _{o}$ Formula (1)

$V_{oy} = V_{o}Sin\alpha _{o}$ Formula (2)

Where:

Vo: Initial velocity in m/s

$\alpha_{o}$ : Initial angle above the horizontal in grades

2) The formula to calculate its velocity at any vertical position(y) is as follows:

$v_{y}^{2} = v_{oy}^{2} -2gy$ Formula (3)

Where:

$v_{f}^{2}$ : Final speed component in vertical direction in m/s

$v_{oy}^{2}$ : Initial speed component in vertical direction in m/s

g: acceleration due to gravity in m/s2

3) The formulas to calculate the projectile velocity components at any time (t) are:

$v_{x} = v_{ox}$ Because the movement is uniform in the x direction (constant speed)

$v_{y} = v_{oy}-g*t$ Formula (4) Because the movement is uniformly accelerated in the y direction

Known information:

We know the following data:

$v_{o} = 65.2 \frac{m}{s}$

$\alpha _{o}$ = 34.5º above the horizontal

$g=9.8 \frac{m}{s^{2}}$

Development of the problem:

Initial speed components(Vox, Voy), (Formula (1), Formula (2)

$v_{ox} = 65.2*Cos34.5 = 53.7\frac{m}{s}$

$v_{oy} = 65.2*Sin34.5 = 36.93\frac{m}{s}$

A) Maximum height (h):

When the projectile reaches its maximum height (h) ,the speed component Vy = 0, then, we replace this value in the Formula (3):

$0=(36.93)^2-2*9.8*h$

$h=\frac{ (36.93)^2}{2*9.8} = 69.58 m$

D) Speed of the projectile 1.50s after firing

We replace t=1.5 s in the formula(4)

$v_{y} = 36.93-9.8*1.5 = 22.23 \frac{m}{s}$

$v_{x} = v_{ox} = 53.7 \frac{m}{s}$

$v = \sqrt{53.7^{2}+22.23^{2}} = 58.12 \frac{m}{s}$ (Speed magnitude)

$\alpha = tan^{-1} (\frac{v_{y}}{v_{x}}) = tan^{-1} (\frac{22.23}{53.7})$

α = 22.49° (Speed direction above the horizontal)

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frozen [14]

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3 years ago