Answer:
116.3 grCO2
Explanation:
1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side
C6H6 +15/2 O2⟶ 6CO2 +3 H2O
2nd - we calculate the limiting reagent
39.2gr C6H6*(240grO2/78grC6H6)=120 grO2
we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent
3rd - we use the limiting reagent to calculate the amount of CO2 in grams
105.7grO2*(264grCO2/240grO2)=116.3 grCO2
F the solubility of a gas in water is 5.0g/L when the pressure of the gas above the water is 2.0 atm, what is the pressure of the gas above the water <span>when the solubility of the gas is 1.0 g/L</span>
Here's how to solve this one.
The formula for solubility is
<span>P1 / P2 = solubility1 / solubility2 </span>
P1=2*1/5= .4 atm
So the correct answer is 0.4 atm.
Answer: I think the answer is C. NaCl and H2O
Explanation: I’m not sure tho
Answer:
Assume that 100 grams of C2H4 is present. This means that there are 85.7 grams of carbon and 14.3 grams of hydrogen.
Convert these weights to moles of each element:
85.7 grams carbon/12 grams per mole = 7 moles of carbon.
14.3 grams hydrogen/1 gram per mole = 14 moles of hydrogen.
Divide by the lowest number of moles to obtain one mole of carbon and two moles of hydrogen.
Since we know that there cannot be a stable CH2 molecule, multiply by two and you have C2H4 which is ethylene - a known molecule.
The secret is to convert the percentages to moles and find the ration of the constituents.
Ba 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² → [Xe]6s²
Ba - 2e⁻ → Ba⁺² [Xe]
