Risks can be changed.<br> True<br> False

why HF (hydrogen fluoride) has higher boiling temperature than HCl (hydrogen chloride), even thought HF has lower molecular weig

bekas [8.4K]

Answer:

Boiling point of HF is higher as compared to HCl because of presence of hydrogen bonding in it.

Explanation:

In HF, intermolecular force of attraction is hydrogen bonding.

Hydrogen bonding is a type of electrostatic force of attraction existing between H atom and electronegative atom.

For a molecule to have hydrogen bonding, H atom must be bonded to electronegative atom, O, N and F.

Hydrogen bonding can be intermolecular and intramolecular.

So, in HF hydrogen bonding present.

In HCl, only van der Waals force exists. van der Waals forces are weak as compared to hydrogen bonding.

Because of presence of hydrogen bonding, HF molecules are held tightly and so requires more heat to boil.

Therefore, boiling point of HF is more as compared to HCl.

5 0

3 years ago

Use the drop-down menus to complete the statements about using OneNote in Outlook meeting requests.

romanna [79]

Answer:

OneNote can be used to share

✔ information

before the meeting.

OneNote integrates seamlessly with Outlook by placing a

✔ link

in the meeting request.

OneNote can also be used to pass information along both

✔ during

and after the meeting.

8 0

2 years ago

Oliver is designing a new children’s slide to increase the speed at which a child can descend. His first design involved steel b

AVprozaik [17]

Answer:

The correct option is;

A) Steel becomes too hot in the Sun and can burn the children

Explanation:

The properties of steel includes;

Low specific heat capacity, high thermal and electrical toughness, high hardness, high tensile strength, high yield strength, appreciable elongation, high fatigue strength, can easily corrode, high malleability and ability to creep

Therefore, due to the low specific heat capacity, which is 0.511 J/(g·°C) and high conductivity of steel which is about 32 W/(m·k), the temperature of the steel can rapidly rise and the hot steel surface can readily conduct the heat, (due to the temperature difference) to other bodies that come in contact

8 0

3 years ago

Which is a better hydraulic cross section for an open channel: one with a small or a large hydraulic radius?

Cerrena [4.2K]

Hydraulic radius is caused by pressurized hydrogen air so that should mean the answer is hydraulic radius

6 0

2 years ago

Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the comp

telo118 [61]

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = $c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}$

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

$-0.40= 1.005(ln T_2 - 5.68697)- 0.5968$

Solving for T2 we have:

$T_2 = 5.8828$

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

$w_in = c_p(T_2 - T_1)+q_out$

$w_in = 1.005(358.8 - 295)+120$

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = $\frac{q_out}{T_1}$

$\frac{120kJ/kg.k}{295K}$

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

3 0

3 years ago

Risks can be changed. True False