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BlackZzzverrR [31]
3 years ago
5

Explain what happens during the chemical reaction CaCO3 -> CaO + CO2

Chemistry
1 answer:
faust18 [17]3 years ago
6 0

check out this article i found it very helpful,

I couldn't find the answer to your question.

Download pdf
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Please help me on this.
Anika [276]
True will end up being the answer
6 0
3 years ago
Read 2 more answers
What is the change of state involved in combustion of an alcohol and it’s entropy ?
Tanya [424]

Answer:

idkwhat is thr qution

Explanation:

4 0
3 years ago
What effect does a catalyst have on a reaccion rate an the products formed in a reaccion
elena-s [515]
A catalyst will speed up the activation energy and therefore speed up the reaction. The products will form fast because of this.
8 0
2 years ago
What sort of evidence would support at least one of the components of the
Yuki888 [10]

Answer:

(not sure) The origin of all cells observed can be traced to previous cells

Explanation:

- not all cells convert energy (?) (this I'm not sure but think so? might be wrong tho)

- not all cells have cell walls e.g. animal cells only have the cell membrane, no cell wall

- Rocks and other nonliving things are composed of cells: non living things are composed of the elements and not cells (unless they were previously alive, then they contain dead cells)

8 0
3 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
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