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nikklg [1K]
1 year ago
15

If a sample of gas is at 672 °C and 3.9 atm, what is the new temperature at 12.2 atm?

Chemistry
1 answer:
Snowcat [4.5K]1 year ago
7 0
In order to calculate the final temperature of the gas, we may apply Charles's law, which states that the pressure and temperature of a fixed amount of gas at constant volume are directly proportional. Mathematically:
P/T = constant
(absolute temperature is used, so T = 672 + 273 = 945 K)
Thus,
3.9 / 945 = 12.2 / T
T = 2,956 K or 2,683 °C
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Be sure to answer all parts. calculate δg o for the reaction between i2(s) and br−(aq). e o cell = −0.54 j/c enter your answer i
denpristay [2]

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

The attached file have the solved problem.

3 0
2 years ago
Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?
kramer

Answer:

Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?

a. Al (s)

b. H2O (l)

c. HCN (g)

d. CH3COOH (l)

e. C2H6 (g)

Explanation:

Entropy is the measure of the degree of disorderness.

In solids, the entropy is very less compared to liquids and gases.

The entropy order is:

solids<liquids<gases

Among the given substances, water in liquid form has a strong intermolecular H-bond.

So, it has also less entropy.

Next acetic acid.

Between the gases, HCN, and ethane, ethane has more entropy due to very weak intermolecular interactions.

HCN has slight H-bonding in IT.

Hence, the entropy order is:

Al(s) < CH3COOH (l) <H2O(l) < HCN(g) < C2H6(g)

7 0
1 year ago
Which statement is TRUE regarding the macroscopic and
damaskus [11]

Answer:

Chemists make observations on the macroscopic a scale that lead to conclusions about microscopic features

Explanation:

Many important chemical observations are made on the macroscopic scale. This is because, many of the scientific equipments available are not presently able to provide direct evidence about microscopic processes. Evidences obtained from macroscopic observations could serve as important insights into the nature of certain microscopic processes.

This is evident in the study of the structure of the atom. Most of the evidences that led to the deduction of the atomic structure were obtained from macroscopic evidence but ultimately provided important information about the microscopic structure of the atom.

7 0
2 years ago
Analysis of the water content of a lake found in the desert showed that it contained 16.6 percent chloride ion, and had a densit
IRISSAK [1]

Answer : The molarity of the chloride ion in the water is, 5.75 M

Explanation :

As we are given that 16.6 % chloride ion that means 16.6 grams of chloride ion present 100 grams of solution.

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}

\text{Volume of solution}=\frac{100g}{1.23g/mL}=81.3mL

Now we have to calculate the molarity of chloride ion.

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of chloride ion}\times 1000}{\text{Molar mass of chloride ion}\times \text{Volume of solution (in mL)}}

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

\text{Molarity}=\frac{16.6g\times 1000}{35.5g/mole\times 81.3mL}=5.75mole/L=5.75M

Thus, the molarity of the chloride ion in the water is, 5.75 M

8 0
2 years ago
If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr
Alinara [238K]

Answer:

The value of  the heat capacity of the Calorimeter  C_c = 54.4 \frac{J}{c}

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water m_w = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q = m_w C_w ΔT + C_c ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +  C_c × 11.24

611.37 =  C_c × 11.24

C_c = 54.4 \frac{J}{c}

This is the value of  the heat capacity of the Calorimeter.

7 0
2 years ago
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