The continuous cycling of carbon in the earyh is known as the carbon cycle.
<h3>What is the carbon cycle?</h3>
The carbon cycle is a cycle which explains the various processes by which carbon is recycled between the atmosphere and the earth.
The constant flow of carbon on earth through organisms and the air is known as the carbon cycle. Plants absorb carbon from the atmosphere in the form of carbon dioxide. They use it for respiration and food production.
Animals consume plants, transferring carbon along the food chain.
Plants and animals respire and release carbon into air. When the animals and plants die, they are eaten by decomposers.
Carbon enters back into the atmosphere in the form of carbondio dioxide.
Therefore, the continuous cycling of carbon in the earth is known as the carbon cycle.
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Answer:
d = 1.85 g/cm³
Explanation:
Given data:
Volume of sulfuric acid = 35.4 mL
Mass of sulfuric acid = 65.14 g
Density of sulfuric acid = ?
Solution:
1 ml = 1cm³
Formula:
d = m/v
d = 65.14 g / 35.2 cm³
d = 1.85 g/cm³
The choices can be found elsewhere and as follows:
A)The reaction requires the collision of three particles with the correct energy and orientation.
<span>B) All reactions that occur in one step are slow. </span>
<span>C)The probability of an effective three-particle collision is low. </span>
<span>D) The transition state is low in energy.
</span>
I think the correct answer from the choices listed above is option C. If the uncatalyzed reaction occurs in a single elementary step, it is a slow reaction because the probability of an effective three-particle collision is low.
You should clean up after every investigation because if you leave a mess, maybe another detective will come in and get lost because of the mess you left.
C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
