Answer:
What can cause excessive pressure on the high side of an active self-contained recovery device? A closed recovery tank inlet valve or excessive air or other non condensables in the recovery tank (either A or B) Portable refillable tanks or containers used to ship recovered refrigerants must meet what standard(s)?
Explanation:
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The answer is 6.1*10^-3 atm.
The pictures and explanations are there.
0.216g of aluminium compound X react with an excess of water water to produce gas. this gas burn completely in O2 to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure
0.108 / n = 24 / 1
n = 0.0045 mole ( CO2 >>0.0045 mole
0.216 - 0.0045 = 0.2115
so Al = 0.2115 / 27 => 0.0078 mole
C = 0.0045 * 1000 => 4.5 and Al = 0.0078 * 1000 = 7.8
Answer:
0.66g of water
Explanation:
Molar heat of vaporization of any substance is defined as the heat necessary to vaporize 1 mole of the substance.
If heat of vaporization of water is 40.79kJ/mol and you add 1.50kJ, the moles you vaporize are:
1.50kJ × (1mol / 40.79kJ) = 0.0368 moles of water.
As molar mass of water is 18.01g/mol, mass of water that can be vaporized are:
0.0368 moles × (18.01g / mol) = <em>0.66g of water</em>
Answer:
<span>23.6
g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from
15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because
it gets used up and only makes 10.7 g of CO2. </span>
Explanation:
1) Balanced chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O
3) molar masses
CH₄: 16.04 g/mol
O₂: 32.0 g/mol
CO₂: 44.01 g/mol
4) Convert the reactant masses to number of moles, using the formula
number of moles = mass in grams / molar mass
CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
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O₂: 15.6 g / 32.0 g/mol = 0.4875 moles
5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂
Which is what the first part of the answer says.
6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.
Which is what the second part of the answer says.
7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.
Which is what the chosen answer says.
8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.
