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3 years ago

To change an object from a liquid to a gas _________ must happen

Chemistry

1 answer:

nignag [31]3 years ago

4 0

Evaporation!
-Melting is a solid to a liquid
-sublimation is directly from a solid to a gas
-condensation gas to liquid

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How do hydrogen ions (h+) and chloride ions (cl–) get into the lumen of the stomach?

user100 [1]

In the stomach cell, the enzyme carbonic anhydrase converts one molecule of carbon dioxide and one molecule of water indirectly into a bicarbonate ion (HCO3-) and a hydrogen ion (H+). In the stomach ion exchange is used to move H+ ions out the cells and into the lumen of the stomach

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3 years ago

A gas system contains 2.00 moles of O2 and CO2 gas, has an initial temperature of 25.0 oC and is under 1.00 atm of pressure. If

Ainat [17]

Answer:

$V_2=52.2L$

Explanation:

Hello there!

In this case, by bearing to to mind the given conditions, it is firstly possible to determine the initial volume of the closed system via the ideal gas equation:

$PV=nRT\\\\V=\frac{2.00mol*0.08206\frac{atm*L}{mol*K}*298.15K}{1.00 atm} \\\\V=48.9L$

Which is V1 in the Charles' law:

$\frac{T_2}{V_2} =\frac{T_1}{V_1}$

And of course, T1 is 298.15 (25+273.15). Therefore, by solving for V2 as the final volume, we obtain:

$V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{48.9L*(45+273.15)K}{(25+273.15)K} \\\\V_2=52.2L$

Best regards!

8 0

3 years ago

Or each set of elements represented in this periodic table outline, identify the principal quantum number, n , and the azimuthal

Tems11 [23]

The answer is potassium. It would be 4, and for neon would be 2. Just total which row of the periodic table you are on. The "L" tells you whether the highest-energy electron is in an "s" orbital (L=0) or a "p" orbital (L=1) or a "d" orbital (L=2) or an "f" orbital (L=3). The way in which these orbitals are filled is: for each of the first three rows (up to argon), two electrons in the "s" orbital are filled first, then 6 electrons in the "p"orbitals. The row where the potassium also starts with filling the "s" orbital at the new "n" level (4) but then goes back to satisfying up the "d" orbitals of n=3 before it seals up the "p"s for n=4.

5 0

3 years ago

Read 2 more answers

Iteams in house made of thermosetting plastics?

Law Incorporation [45]

Thermoplastics and thermosetting polymers Examples include: polyethylene (PS) and polyvinyl choline (PVC). Common thermoplastics range from 20,000 to 50,000 amu, while thermosets are assumed to have infinite molecular weight.

3 0

3 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

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3 years ago