All categories

3 years ago

The freezing point of a substance is the at which it freezes.

Chemistry

2 answers:

NeX [460]3 years ago

8 0

Answer:

reff lagay mo yung ilalagay mo

pshichka [43]3 years ago

3 0

Answer:

Freezing point is the temperature at which a liquid becomes a solid at normal atmospheric pressure. Alternatively, a melting point is the temperature at which a solid becomes a liquid at normal atmospheric pressure.

You might be interested in

Which equation represents a neutralization reaction?

abruzzese [7]

C. HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O

8 0

4 years ago

12. Which atoms or group of atoms make up the functional group of ethanol (CH3CH2OH)?

Montano1993 [528]

Answer:

Option A. the hydroxyl group (-OH)

Explanation:

Ethanol, CH₃CH₂OH belongs to the class of organic compound called alkanol.

They have general formula as R–OH

Where

R => is an alkyl group

OH => is the hydroxyl group

The hydroxyl group (OH) is the functional group of the alkanol (alcohol)

8 0

3 years ago

Please help soon!!! About Chemistry and noble gas notations.

Svetach [21]

Answer:

play video games all days

Explanation:

ggggggggghahahahahahaha

8 0

3 years ago

What is true about all matter?

Ipatiy [6.2K]

Alll are correct. Matter is literally everywhere

8 0

3 years ago

The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen gas. What was the pressure in the

Ray Of Light [21]

Answer:

$\large \boxed{\text{103 kPa}}$

Explanation:

We can use the Ideal Gas Law — pV = nRT

Data:

V = 66.8 L

m = 77.8 g

T = 25 °C

Calculations:

(a) Moles of N₂

$\text{Moles of N}_{2} = \text{77.8 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{28.01 g N}_{2}} = \text{2.778 mol N}_{2}$

(b) Convert the temperature to kelvins

T = (25 + 273.15) K = 298.15 K

(c) Calculate the pressure

$\begin{array}{rcl}pV & =& nRT\\p \times \text{66.8 L} & = & \text{2.778 mol} \times \text{8.314 kPa$\cdot$ L$\cdot$K$^{-1}$mol$^{-1}\times$ 298.15 K}\\66.8p & = & \text{6886 kPa}\\p & = & \textbf{103 kPa}\end{array}\\\text{The pressure in the bag is $\large \boxed{\textbf{103 kPa}}$}$

3 0

3 years ago