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3 years ago

What is the pressure, in atmospheres, of 6.0 grams of H2 gas if it has a volume of 73 liters when the temperature is 25 C?

1 answer:

5 0

PV = nRT (where P = pressure; V = volume; n = number of moles; R = gas constant; T = Temperature)

Moles of H₂ = mass of H₂ ÷ molar mass of H₂
= 6.0 g ÷ 2 g/mol
= 3.0 mol

To get the answer in atmospheres, let's convert C to K
25 °C ≡ 298 K

Now, based on the formula above
P = (nRT) ÷ V

P = (3.0 mol × (0.082 L · atm/mol · K) × 298 K) ÷ 73 L
≈ 1 .004 atm

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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:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

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3 years ago

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3 years ago

Read 2 more answers

) palmitic acid is a 16 carbon acid. in a balanced equation, the products of the saponification of glyceryl tripalmitate (tripal

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Answer: The products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.

Explanation:

A chemical reaction in which triglycerides react with sodium hydroxide and leads to the formation of one molecule of glycerol and three molecules of a salt of fatty acid is known as saponification.

For example, when tripalmitin reacts with sodium hydroxide then it leads to the formation of one molecule of glycerol and three molecules of sodium salt of palmitic acid.

The reaction equation is as follows.

$C_{51}H_{98}O_{6} + NaOH \rightarrow C_{3}H_{8}O_{3} + 3C_{16}H_{31}O_{2}Na^{+}$

Thus, we can conclude that the products of the saponification of glyceryl tripalmitate (tripalmitin) are one molecule of glycerol and three molecules of sodium salt of palmitic acid.

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4 years ago

0.20 L of HNO3 is titrated to equivalence using 0.12 L of 0.2 M NaOH. What is the concentration of the HNO3?

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