What is the volume of 0.200 mol of an ideal gas at 200. kPa and 400. K?

How would life be different if there were 10 hours in a day

Anna71 [15]

Answer:

If there was only 10 hours in a day, including the night time it would be complexly new environment. Say you are up for 7 hours and sleep 3, in those 7 hours you would go to work for probably 4 hours, come home and do stuff for 3 then go to bed and do it all over again.

8 0

3 years ago

Which element is a noble gas?<br> A. W<br> B. Ar<br> C. N<br> D. Er

inessss [21]

His the answer I hope it helps

3 0

3 years ago

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$

Thereby, the magnitude and direction of work turn out:

$W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ$

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0

3 years ago

Which of the following statements is FALSE?

vovikov84 [41]

Answer: ITS FALSE

ExplaC6H12 + 9 O2 → 6 H2O + 6 CO2 is a single replacement reaction

nation:

3 0

2 years ago

Name each of the following species for the following acid-base reactions. (The equilibrium lies to the right in each case, i.e.,

DaniilM [7]

Answer: a) $H_3O^++CH_3O^-\rightleftharpoons CH_3OH+H_2O$

acid : hydronium ion

base : methoxide ion

conjugate acid : methanol

conjugate base: water

b) $CH_3CH_2O^-+HCl\rightleftharpoons CH_3CH_2OH+Cl^-$

acid : hydrogen chloride

base : ethoxide ion

conjugate acid : ethanol

conjugate base: chloride ion

c) $NH_2^-+CH_3OH\rightleftharpoons NH_3+CH_3O^-$

acid : methanol

base : amide ion

conjugate acid : ammonia

conjugate base: methoxide ion

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

The species accepting a proton is considered as a base and after accepting a proton, it forms a conjugate acid.

The species losing a proton is considered as an acid and after loosing a proton, it forms a conjugate base

For the given chemical equation:

a) $H_3O^++CH_3O^-\rightleftharpoons CH_3OH+H_2O$

acid : hydronium ion

base : methoxide ion

conjugate acid : methanol

conjugate base: water

b) $CH_3CH_2O^-+HCl\rightleftharpoons CH_3CH_2OH+Cl^-$

acid : hydrogen chloride

base : ethoxide ion

conjugate acid : ethanol

conjugate base: chloride ion

c) $NH_2^-+CH_3OH\rightleftharpoons NH_3+CH_3O^-$

acid : methanol

base : amide ion

conjugate acid : ammonia

conjugate base: methoxide ion

.

3 0

2 years ago