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gtnhenbr [62]
2 years ago
12

What volume does .0685 mol of gas occupy at STP? 1mole =22.4

Chemistry
2 answers:
SashulF [63]2 years ago
6 0

Answer : The volume of gas at STP is, 1.53 liter

Solution : Given,

Moles of a gas = 0.0685 mole

As we know that at STP,

1 mole of gas contains 22.4 liter volume of gas

As, 1 mole of gas contains 22.4 liter volume of gas

So, 0.0685 mole of gas contains 0.0685\times 22.4=1.53L volume of gas

Therefore, the volume of gas at STP is, 1.53 liter

GuDViN [60]2 years ago
4 0
V = nRT/P
V = 0.685 mol*(.0821 L*atm/K*mol)*273 K/1 atm
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satela [25.4K]

Answer:

cell wall

Explanation:

plants are the only cells that have cell walls. no other type of organism have a cell wall

6 0
2 years ago
8. The standard enthalpy of formation of RbF(s) is –557.7kJ/mol and the standard enthalpy of formation of RbF(aq, 1 m) is –583.8
garri49 [273]

Explanation:

Given

The enthalpy of formation of RbF (s) is –557.7kJ/mol

The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol

The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)

= -583.8 - (-557.7)  kJ/mol

= -26.1 kJ/mol

The enthalpy is negative which means that the temperature will rise when RbF is dissolved.

3 0
3 years ago
In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2
valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

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mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

6 0
3 years ago
During combustion, methane yields carbon dioxide and water. The unbalanced equation for this reaction is:CH4(g)+O2(g) → CO2(g)+
almond37 [142]

Answer:

CH₄(g)  +  2O₂(g) → CO₂(g)  +  2H₂O(l)

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2

Explanation:

CH4(g)+O2(g) → CO2(g)+ H2O(l)

To balance a chemical equation, you must have the same mole of each element in both sides of the reaction (reactant side and product side)

CH₄(g)  +  2O₂(g) → CO₂(g)  +  2H₂O(l)

2 C

8 H

8 O

In  both side.

It takes 1 mole of methane to react with 2 mole of oxygen in order to produce 1 mol of dioxide and 2 mole of water.

Mole ratios for the balanced equation be:

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2  - We should compare each compound

1 mol methane → 2 mole of oxygen

2 mole of oxygen → 1 mol of methane

2 mole of oxygen → 1 mol of dioxide

1 mole of dioxide → 1 mol of Methane

2 mole of water → 2 mole of oxygen (the same as opposite)

1 mol methane →  2 mole of water

2 mol of water → 1 mol of methane

4 0
3 years ago
That's just the tip of the iceberg" is a popular expression you may have heard. It means that what you can see is only a small p
puteri [66]
Answer:

B 1.23 g/cc

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Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.

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B. 1.23 g/cc
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D. 4.14 g/cc
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5 0
3 years ago
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