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velikii [3]
3 years ago
12

How does the burning of fossil fuels increase the greenhouse effect answers?

Chemistry
1 answer:
miskamm [114]3 years ago
8 0
The burning of fossil fuels like coal, natural gas etc... leads to the emmission of carbondioxide which is a green house gas and leads to the increase of heat in the atmosphere. This results in greenhouse effect.
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FIRST TO ANSWER CORRECTLY GETS BRAINELSIT! pls asap :)
zmey [24]
The protons! Is it the answer
4 0
3 years ago
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A gold nugget has a density of 38.6g/cm3 and a mass of 270.2. what is its volume?​
aivan3 [116]

Density is given by the equation D=m/V, were D is density, m is mass in grams, and V is volume in cubic centimeters.


In this problem, we have density and we have mass so we can plug into the equation and solve for V.

38.6=270.2/V

<em>*Multiply both sides by V*</em>

38.6V=270.2

<em>*Divide both sides by 38.6*</em>

V=7


The volume of the gold nugget is 7cm3.


Hope this helps!!

5 0
3 years ago
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Bismuth oxide reacts with carbon to form bismuth metal: bi2o3(s) + 3c(s) → 2bi(s) + 3co(g) when 689 g of bi2o3 reacts with exces
ExtremeBDS [4]
Using the answer from the first part, we know that 2.957 moles of bismuth have formed. Moreover, the molar ratio between bismuth and carbon monoxide is:

2 : 3

Using the method of ratios,

2 : 3
2.957 : CO

CO = (3 * 2.957) / 2
CO = 4.4355

4.436 moles of carbon monoxide will be formed
8 0
3 years ago
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The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c
crimeas [40]

Answer : The rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 785.0K = ?

Ea = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 785.0K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]

K_2=1.45\times 10^{-2}s^{-1}

Therefore, the rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

7 0
3 years ago
Assuming that gasoline is 100% isooctane, that isooctane burns to produce only CO2CO2 and H2OH2O, and that the density of isooct
Aleksandr [31]

Answer:

1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane

Explanation:

Let's state the combustion reaction:

C₈H₁₈  +  25/2O₂  →   8CO₂  +  9H₂O

Let's calculate the mass of isooctane that reacts.

Density = Mass / Volume

Density . Volume = Mass

First of all, let's convert the volume in L to mL, so we can use density.

4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL

0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g

This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction

3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol

Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide

Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8)  = 2.55×10¹² moles of CO₂

Now, we can determine the mass of produced CO₂ by multipling:

moles . molar mass

2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂

If we convert to kg  1.12×10¹⁴ g / 1000 =  1.12×10¹¹ kg

6 0
3 years ago
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