All categories

3 years ago

A real gas behaves least like an ideal gas under the conditions of

1 answer:

5 0

The correct answer is option 3. A real gas behaves least like an ideal gas under the conditions of high temperature and low pressure. At this temperature and pressure, the molecules are close to each other and collisions or interactions are very likely to happen which is not an ideal gas.

You might be interested in

Select the keyword or phrase that will best complete each sentence.

pishuonlain [190]

Answer:

1) acetylide

2) enol

3) aldehydes

4) tautomers

5) alkynes

6) Hydroboration

7) Keto

8) methyl ketones

Explanation:

Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.

Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.

The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.

3 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

In which of these situations is convection most likely the main form of heat transfer?

lorasvet [3.4K]

In the first situation: the mechanism of covection is the main form of heat transfer when warm air from a heater moves around and upward.

In the case of the metal pan the mechanism of heat transfer is conduction.

In the case of sunburn the mechanism is radiation.

In the case of an ice cube melting in a hand, conduction is the most important mechanism.

5 0

3 years ago

Read 2 more answers

The rate constant for this second‑order reaction is 0.610 M − 1 ⋅ s − 1 0.610 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶product

Darya [45]

Answer: It takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.

Explanation:

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$