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3 years ago

Which choice is an example of an endothermic process?

Chemistry

1 answer:

77julia77 [94]3 years ago

7 0

Answer: conversion of ice to steam

Explanation: Endothermic process is one in which energy is absorbed by the system.

Conversion of ice to steam is change of solid phase to gaseous phase, thus energy is required to break the strong inter molecular forces of attraction in solids to convert it into gaseous phase.

Conversion of steam to ice, conversion of steam to water and conversion of water to ice releases energy and are examples of exothermic processes.

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What type of reaction is displayed below?

mr Goodwill [35]

Answer:

double replacement

Explanation:

The reaction shown is a double replacement reaction.

It is also known as double decomposition or metathesis reaction.

In such a reaction, there is an actual exchange of partners to form new compounds.

One of the following is the driving force for such reaction:

formation of an insoluble compound or precipitate
formation of water or any other non-ionizing compound
liberation of a gaseous product.

3 0

3 years ago

Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze

Rashid [163]

Answer:

a. Δ $H^0_{rxn} = -108.0\frac{kJ}{mol}$

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate Δ $H^0_{rxn}$ , for the gas-phase reaction ) using the Hess's Law.

Δ $H^0_{rxn}$ = $E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}$

Given from the question, the table below shows the corresponding Δ $H^0_{t}(kJ/mol)$ for each compound.

Compound $H^0_{t}(kJ/mol)$

Liquid EO -77.4

$CH_4_(g_)$ -74.9

$CO_(g_)$ -110.5

If we incorporate our data into the above previous equation; we have:

Δ $H^0_{rxn}$ = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

= $-108.0 \frac{kJ}{mol}$

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

$T_{initial}$ = 93.0°C &

the enthalpy of vaporization (Δ $H^0_{vap}$ ) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as = $\frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}$

Let's not forget as well, that Δ $H^0_{vap}$ = $\frac{q}{mass}$

If we substitute Δ $H^0_{vap}$ for $\frac{q}{mass}$ in the above equation, we have;

specific heat capacity (c) = $\frac{deltaH^0_{vap}}{T_{final}-T_{initial}}$

Making ( $T_{final}- T_{initial}$ ) the subject of the formula; we have:

$T_{final}- T_{initial}$ = $\frac{delat H^0_{vap}}{specificheat capacity}$

$(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}$

$T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C$

= 227.76°C +93.0°C

= 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C

7 0

4 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .

Or, 5.0× $10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0× $10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00× $10^{-13}$ ) $^{1/2}$ = 7.071× $10^{-7}$ g/mL.

Thus the concentration of Br $^{-1}$ or HBr is 7.071× $10^{-7}$ g/mL.

4 0

4 years ago

Consider the titration of sulfuric acid with sodium hydroxide. What volume (mL) of a 2.658M NaOH solution is required to fully t

KatRina [158]

The volume of base required to completely neutralize the acid is 3.2 mL of NaOH.

The equation of the reaction is;

2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)

From the question;

Concentration of acid CA = 0.426M

Concentration of base CB = 2.658M

Volume of acid VA = 10.00mL

Volume of base VB = ?

Number of moles of acid NA = 1

Number of moles of base NB = 2

Using the relation;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

VB = CAVANB/CBNA

VB = 0.426M × 10.00mL × 2/ 2.658M × 1

VB = 3.2 mL

Learn more: brainly.com/question/6111443

8 0

3 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