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sdas [7]
3 years ago
10

How many moles of oxygen will occupy a volume of 2.50 liters?

Chemistry
1 answer:
loris [4]3 years ago
7 0
Data:
n (number of mols) = ? 
V (volume) = 2.50 Liters 
If: 
1 L → 1000 g
2.50 L → y

y = 1000*2.50 = 2500 g

Therefore:
m (mass) = 2500 g

Now:
Molar Mass (MM) of oxygen = 16 g/mol

Formula:
n =  \frac{m}{MM}

Solving:
n = \frac{m}{MM}
n = \frac{2500\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol}
\boxed{\boxed{n = 156.25\:mols}}\end{array}}\qquad\quad\checkmark




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Complete combustion of 7.50 g of a hydrocarbon produced 23.1 g of CO2 and 10.6 g of H2O. What is the empirical formula for the h
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A CH compound is combusted to produce CO2 and H2O 
CnHm + O2 -----> CO2 + H2O 
Mass of CO2 = 23.1g
 Mass of H2O = 10.6g
 Calculate by mass of the compounds
 For Carbon C, divide by molecular weight of CO2 and multiply with Carbon
molecular weight. So C in grams = 23.1 x (12.01 / 44.01) = 6.3 g C
 For Hydrogen H, divide by molecular weight of H2O and multiply with Hydrogen molecular weight. So H in grams = 10.6 x (2.01 / 18.01) = 0.53 g C
= 1.18 of H
 Calculate the moles for C and H
 6.3 grams of C x (1 mole/12.01 g C) = 0.524 moles of C
 1.18 grams of H x (1 mole/1.008 g H) = 1.17 moles of H
 Divides by both mole entities with smallest
 C = 0.524 / 0.524 = 1 x 4 = 4
 H = 1.17 / 0.524 = 2.23 x 4 = 10
 The empirical formula is C4H10.
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Answer:

what kind of chemistry is it going to be?

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A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f
a_sh-v [17]

Answer:

m_{H_2O}=39.0g

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

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