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ozzi
2 years ago
7

Balance the equation ___ CaCO3 -> ____ CaO + ____ CO2

Chemistry
1 answer:
Degger [83]2 years ago
3 0

Answer:

CaCO3 -> CaO + CO2

Explanation:

Woahhhh, did you balance it yourself just then?

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Purest form of carbon is
vladimir2022 [97]

Answer:

Obtención. El carbono se encuentra - frecuentemente muy puro - en la naturaleza, en estado elemental, en las formas alotrópicas diamante y grafito. El material natural más rico en carbono es el carbón (del cual existen algunas variedades). Grafito: Se encuentra en algunos yacimientos naturales muy puro.

4 0
3 years ago
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For the Earth's atmosphere, section number two in the pie graph BEST represents the percentage of A) argon B) oxygen C) nitrogen
larisa86 [58]

-<u><em>Oxygen</em></u>

According to Google these are the percentages of the <em>Earths Atmosphere</em>

<em>1</em> 78% - Nitrogen

<u>2</u> 21% - Oxygen

<em>3</em> 0.9% - Argon

<em>4 </em>0.3 - Carbon Dioxide with very small percentage of other elements.

3 0
3 years ago
Nnbvcvbnm,.,mnbvcxcvbjklkjhvbn
pav-90 [236]
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3 0
3 years ago
Combining hydrogen and oxygen to make water Group of answer choices releases a lot of energy. consumes a lot of energy. releases
Firdavs [7]

Answer:

Releases a lot of energy. When hydrogen a highly reactive element reacts with oxygen, it does so using immense force, when the bonds break high amounts of emery is released.

Explanation:

5 0
3 years ago
A 75.0-milliliter lightbulb is filled with neon. There are 7.16 × 10-4 moles of gas in it, and the absolute pressure is 116.8 ki
Fantom [35]

Answer:- 1467 K

Solution:- It asks to calculate the kelvin temperature of the light bulb. Looking at the given info, it is based on ideal gas law equation, PV=nRT.

Given: n=7.16*10^-^4moles

V = 75.0 mL = 0.0750 L

P = 116.8 kPa

We know that, 101.325 kPa = 1 atm

So, 116.8kPa(\frac{1aym}{101.325kPa})

= 1.15 atm

R is universal gas constant and it's value is 0.0821\frac{atm.L}{mol.K} .

T = ?

Let's plug in the values in the equation and solve it for T.

1.15(0.0750)=7.16*10^-^4*0.0821(T)

0.08625 = 0.00005878(T)

T=\frac{0.08625}{0.00005878}

T = 1467 K

So, the temperature of the light bulb would be 1467 K.

6 0
2 years ago
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