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Rama09 [41]
3 years ago
5

What is the difference between a volume that is delivered and a volume that is contained?

Chemistry
1 answer:
nikdorinn [45]3 years ago
3 0

Answer:

When the graduation line denotes the volume contained in the calibrated vessel, the ware is marked “TC”. When the graduation line indicates the volume delivered from the vessel, the ware is marked “TD”.

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Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
In the equilibrium constant expression for the reaction below what is the correct exponent for N2O4?
irga5000 [103]
As we have the balanced reaction equation is:

N2O4 (g) ↔ 2NO2(g)

from this balanced equation, we can get the equilibrium constant expression

KC = [NO2]^2[N2O4]^1

from this expression, we can see that [NO2 ] is with 2 exponent of  the stoichiometric and we can see that from the balanced equation as NO2
is 2NO2 in the balanced equation.

and [N2O4] is with 1 exponent of the stoichiometric and we can see that from the balanced equation as N2O4 is 1 N2O4 in the balanced equation. 

∴ the correct exponent for N2O4 in the equilibrium constant expression is 1 
7 0
3 years ago
in rutherford's gold foil experiment a very small number of alpha particles were deflected. what about the structure of the atom
zvonat [6]

Answer:The atom being mostly empty space. A small number of alpha particles were deflected by large angles (> 4°) as they passed through the foil. There is a concentration of positive charge in the atom. Like charges repel, so the positive alpha particles were being repelled by positive charge

8 0
3 years ago
Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---&gt; C2H2(g) ΔH° = ??? kJ Given the following thermochemical equation
nordsb [41]

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

C_2H_2(g) + \frac{5}{2}O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_{1} = -1,123 kJ...[1]

C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_{2} = -340 kJ..[2]

H_2(g) + \frac{1}{2}O_2(g)\rightarrow H_2O(l) ,\Delta H^o_{3} = -211 kJ..[3]

2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_{4} =?..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}

\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)

\Delta H^o_{4}=232 kJ

The enthalpy for given reaction is 232 kilo Joules.

5 0
3 years ago
A scale model of the solar system
Ket [755]
There you go BRAINLIEST would be appreciated

3 0
3 years ago
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