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Sonja [21]
3 years ago
12

2-An analysis that establishes the chemical identity of the species

Chemistry
2 answers:
In-s [12.5K]3 years ago
8 0
B)Quantitative Synthesis
saw5 [17]3 years ago
7 0

Answer:

C Qualitative Analysis.

Explanation:

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What is the lewis structure for HCL + Na?
saul85 [17]

Answer:

Explanation: that is .....

3 0
3 years ago
Give the name and symbol of your favorite element and explain why it your favorite element.​
Salsk061 [2.6K]

Answer:

The most important elements that we use in everyday life include carbon, hydrogen, oxygen, with smaller amounts of things like chlorine, sulfur, calcium, iron, phosphorus,nitrogen, sodium, and potassium. Apart from these, other elements include magnesium, zinc, neon, and helium are also in our daily existence.

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6 0
2 years ago
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0
3 years ago
A balloon is filled with 0.250 mole of air at 35°C. If the volume of the balloon is 6.23 liters, what is the absolute pressure
lord [1]

Answer is:  the absolute pressure of the air in the balloon is 1.015 atm (102.84 kPa).

n = 0.250 mol; amount of substance.

V = 6.23 L; volume of the balloon.

T = 35°C = 308.15 K; temperature.

R = 0.08206 L·atm/mol·K, universal gas constant.

Ideal gas law: p·V = n·R·T.

p = n·R·T / V.

p = 0.250 mol · 0.08206 L·atm/mol·K · 308.15 K / 6.23 L.

p = 1.015 atm; presure of the air.

6 0
3 years ago
Calculate the internal energy of 2 moles of argon gas (assuming ideal behavior) at 298 K. Suggest two ways to increase its inter
Dominik [7]
I hope this helps you.

5 0
3 years ago
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