Answer:
mCO2= 49.6932 kgCO2
Explanation:
Hello! Let's solve this!
First we propose the balanced equation C3H8 + 5O2 ---> 3CO2 + 4H2O
We see that each mole of C3H8 (propane) we get 3 moles of CO2
From the propane volume we can obtain the grams of propane used.
molpropane = 26.5L * (1000mL / 1L) * (0.621g / 1mL) * (1mol / 44g) = 374.01mol propane
mCO2 = 374.01molC3H8 * (3molCO2 / 1molC3H8) * (44gCO2 / 1molCO2) = 49369.32g * (1kg / 1000g) = 49.6932 kgCO2
mCO2= 49.6932 kgCO2
The answer is B, early in the morning water covered areas (lakes,ponds,puddles,etc.) will vaporize a little bit because of the heat from the sun and it will continue all day, vapors rise towards the atmosphere and since it's a lot cooler there it will condense into a cloud which is full of tiny frozen water particles. Hope this helps <span />
<span>The equation that represents the process of photosynthesis
is: </span>
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</span>
<span>6CO2+12H2O+light->C6H12O6+6O2+6H2O</span>
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<span>Photosynthesis is the
process in plants to make their food. This involves the use carbon dioxide to
react with water and make sugar or glucose as the main product and oxygen as a
by-product. Since we are not given the mass of CO2 in this problem, we assume that we have 1 g of CO2 available. We calculate as follows:</span>
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</span>
<span>1 g CO2 ( 1 mol CO2 / 44.01 g CO2 ) ( 12 mol H2O / 6 mol CO2 ) ( 18.02 g / 1 mol ) = 0.82 g of H2O is needed</span>
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</span>
However, if the amount given of CO2 is not one gram, then you can simply change the starting value in the calculation and solve for the mass of water needed.
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</span>
Answer:
I don't understand what you are asking
