Ask question

Ask question

All categories

2 months ago

15

In the following take CV = 20.8 and CP = 29.1 J⋅mol−1⋅°C−1 for nitrogen gas: (a) Three moles of nitrogen at 30°C, contained in a

rigid vessel, is heated to 250°C. How much heat is required if the vessel has a negligible heat capacity? If the vessel weighs 100 kg and has a heat capacity of 0.5 kJ⋅kg−1⋅°C−1, how much heat is required? (b) Four moles of nitrogen at 200°C is contained in a piston/cylinder arrangement. How much heat must be extracted from this system, which is kept at constant pressure, to cool it to 40°C if the heat capacity of the piston and cylinder is neglected?

1 answer:

eduard [2.7K]2 months ago

4 0

Answer:

$224 \times 13313\frac{.131?}{?244}$

You might be interested in

A chalk company gets an order for 3000 boxes of chalk at the beginning of the school year, the largest order they have ever rece

VMariaS [2998]

The chalk production's efficiency is noted to be 82 %

To determine this efficiency, we utilize the formula,

% Efficiency $=\frac{Actual yield (g)}{Theoretical yield (g)} *100$

Finding the actual yield based on % efficiency and the theoretical yield:

82 % $=\frac{Actual yield}{400,000 g} * 100$

Actual yield = $\frac{82 *400000 g}{100} = 32,800 g$

Each box contains 145 grams of chalk

Total boxes = $32800 g *\frac{1 box}{145 g} = 2262 boxes$

The chalk box company aims to produce 3000 boxes. However, with 82 % efficiency, they can only produce 2262 boxes, thus falling short of their target.

8 0

3 months ago

In a hospital laboratory, a 10.0 mL sample of gastric juice (predominantly HCl), obtained several hours after a meal, was titrat

alisha [2963]

Answer: 1.14

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

To find the molarity of the acid, we will utilize the equation derived from the neutralization process:

$n_1M_1V_1=n_2M_2V_2$

where

are the n-factor, molarity, and volume for the acid and

represent the n-factor, molarity, and volume for NaOH.

We know that: $n_1,M_1\text{ and }V_1$ $HCl$

By substituting the known values into the equation, we get:

$n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL$

To determine the pH of gastric juice:

The molarity amounts to = 0.072

$1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M$

Thus, the pH level of the gastric juice is 1.14

3 0

3 months ago

Based on the crystal-field strengths cl- < f- < h2o < nh3 < h2nc2h4nh2, which octahedral ti (iii) complex below has

eduard [2782]

<span>According to crystal field strength, the Cl ligand results in the longest d-d transition when coordinated with Ti(III) due to its classification as a weak field ligand that causes minimal d orbital splitting.</span>

8 0

2 months ago

Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul

Anarel [2989]

Answer:

$n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}$

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

$n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}$

Where

$n$ is the number of moles of hydrogen

$n$ is the mass of hydrogen

$\rho$ is the density of hydrogen

6 0

4 months ago

Read 2 more answers

he population of the Earth is roughly eight billion people. If all free electrons contained in this extension cord are evenly sp

eduard [2782]

1) Drift velocity: $3.32\cdot 10^{-4}m/s$

2. $5.6\cdot 10^{13}$ electrons per individual

Explanation:

1)

In a conducting material with an electric current, the drift velocity of electrons can be calculated using this equation:

$v_d=\frac{I}{neA}$

where

I stands for current

n represents the density of free electrons

$e=1.6\cdot 10^{-19}C$ indicates the charge of an electron

A signifies the wire's cross-sectional area

The wire's cross-sectional area can be determined as

$A=\pi r^2$

where r denotes the wire's radius. Thus, the equation transforms to

$v_d=\frac{I}{ne\pi r^2}$

In this scenario, we have:

I = 8.0 A as the current

$8.5\cdot 10^{28} m^{-3}$ indicates the free electron concentration

d = 1.5 mm is the diameter, making the radius

r = 1.5/2 = 0.75 mm = $0.75\cdot 10^{-3}m$

So, the resulting drift velocity is:

$v_d=\frac{8.0}{(8.5\cdot 10^{28})(1.6\cdot 10^{-19})\pi(0.75\cdot 10^{-3})^2}=3.32\cdot 10^{-4}m/s$

2)

The entire length of the cord is

L = 3.00 m

And the cross-sectional area is

$A=\pi r^2=\pi (0.75\cdot 10^{-3})^2=1.77\cdot 10^{-6} m^2$

Consequently, the volume of the cord is

$V=AL$ (1)

The number of electrons per unit volume is $n$ , thus the total electrons in this cord would be

$N=nV=nAL=(8.5\cdot 10^{28})(1.77\cdot 10^{-6})(3.0)=4.5\cdot 10^{23}$

Overall, the Earth's population rounds to 8 billion individuals, equating to

$N'=8\cdot 10^9$

Hence, the number of electrons distributed to each person is:

$N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}$

7 0

3 months ago