2H2(g) + O2(g) → 2H2O(1) 0 260 g 0.2068 0.180 g 2008
When 45.0 g of CH4 reacts with excess O2, the actual yield of CO2 is 118 g. What is the percent yield? CHA(g) + 2O2(g) - CO2(g) + 2H2O(g) 73.6% 67.9% 95.2% 86.4%
For the reaction: 2503(g) + 790 kcal - 25(s) + 3O2(g), how many kcal are needed to form 1.5 moles O2(g)? 790 kcal 395 kcal 2370 kcal 411 kcal
When 3 moles of Ny are mixed with 5 moles of H2 the limiting reactant is N2(g) + 3H2(g) - 2NH3(g) H2 NH3 ОООО H20 O N₂
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Answer:</h2><h2>
The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = 19.4 %</h2>
Explanation:
Average electricity consumed per month = 900 kWh
The family cools their house for three months during the summer with two window-unit air conditioners
The power consumed by one window-unit air conditioners = 350 kWh
The power consumed by two window-unit air conditioners = 350(2) = 700 kWh
Power consumed for two air conditioners for the three summer months = 700 (3) = 2100 kWh
Total power consumed for 1 year = 900 (12) = 10800kWh
The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = 
= 19.4 %
Answer:
345.89 g/mol
Explanation:
To find the molar mass, find the atomic mass of all the elements from a periodic table.
Cs - 132.91 × 2 = 265.82
S - 32.07
O - 16.00 × 3 = 48.00
Now add them all together.
265.82 + 32.07 + 48.00 = 345.89 g/mol
Hope that helps.
Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.
Explanation:
- d = m/V, where d is the density, m is the mass and V is the volume.
- We have the mass m = 0.50 g, so we must get the volume V.
- To get the volume of a gas, we apply the general gas law PV = nRT
P is the pressure in atm (P = 1.5 atm)
V is the volume in L (V = ??? L)
n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).
- Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.
- Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.
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