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3 years ago

How many moles of gas are contained in a 30.0 L tube with a temperature of 300. K and pressure at 200. atm ?

Chemistry

1 answer:

Naddik [55]3 years ago

5 0

Answer:

Yes, 244 moles

Explanation:

PV = nRT => n = PV/RT = (200atm)(30.0L)/(0.08206Latm/molK)(300K) = 243.7mol ≅ 244mol

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Hi! I'm on the last question of my homework and I'm stuck on it and was wondering if anyone could help me

Semmy [17]

I am not sure but this is what I think

This question uses the formula connecting mass, density and volume
Which is Density= Mass/Volume

Convert the mass in g
92.5 kg = 92,500g

7.87g/ml = 92,500/ Volume
Volume= 92,500/7.87
= 11,753.5 ml

Now since we have to give the answer in liters we can just divide by 1000 and get
11.75 litres

5 0

2 years ago

The balanced chemical equation for the reaction between sodium chloride and silver nitrate is:

bonufazy [111]

Answer:

We can Interprete it as 1mole of Sodium Chloride and 1mole of Silver Nitrate React to Produce

1Mole of Silver Chloride and 1Mole of Sodium Nitrate

7 0

2 years ago

Read 2 more answers

Estimate the ligand fi eld splitting for (a) [CrCl6]3 (max 740 nm), (b) [Cr(NH3)6]3 (max 460 nm), and (c) [Cr(OH2)6]3 (max 575 n

igor_vitrenko [27]

Answer:

NH3>H2O>Cl-

Explanation:

The given wavelengths of maximum absorption for each complex can be used to estimate the magnitude of field splitting of the respective ligands as shown in the image attached. The field splitting is reported in the unit kilojoule per mole (KJmol-1).

It can be seen from the calculation in the image attached that ammonia shows the highest crystal field splitting followed by water and lastly the chloride anion. This corresponds to the respective positions of these species in the spectrochemical series. Water and the chloride ion are weak field ligands.

4 0

3 years ago

Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)

Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation Cr(s) → Cr3+ + 3e-

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the half-reactions

2(Cr → Cr3+ + 3e-) E° ox = 0.74 V

3(Ni2+ +2e- → Ni) E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox = -0.25 + 0.74 = 0.49

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution) MnO4- +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e- (oxidation) Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the half-reactions

MnO4- +8H+ +10-e- ⇔ Mn2+ +4H2O E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e- E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox = 1.51 + 2.12 = 3.63 V

E = E ° − 0.0257 V /n * ln Q = E ° − 0.0257 V /n *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0

3 years ago

Q1.

valentinak56 [21]

Answer:

4.49dm3

Explanation:

2NH4Cl + Ca(OH)2 —> CaCl2 + 2NH3 + 2H2O

First, we need to convert 10g of ammonium chloride to mole. This is illustrated below:

Molar Mass of NH4Cl = 14 + (4x1) + 35.5 = 53.5g/mol

Mass of NH4Cl = 10g

Number of mole = Mass /Molar Mass

Number of mole of NH4Cl = 10/53.5 = 0.187mol

From the equation,

2moles of NH4Cl produced 2 moles of NH3.

Therefore, 0.187mol of NH4Cl will also produce 0.187mol of NH3

Now we can obtain the volume of NH3 produced by doing the following:

1mole of any gas occupy 24dm3

Therefore, 0.187mol of NH3 will occupy = 0.187 x 24 = 4.49dm3

4 0

3 years ago