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2 years ago

What temperature does ice freeze

Chemistry

2 answers:

Gwar [14]2 years ago

8 0

Answer:

32°F

Explanation:

Water freezes at 32°F.

Vlad1618 [11]2 years ago

6 0

Answer:

At 0 celcius or 32 Fahrenheit

Explanation:

when you want to freeze a substance like water it has to be at that temperature 0 Celsius or 32 Fahrenheit. Also it's not at exact, for example you put the water in 0 celcius and it freeze ni wait a few minutes and it would be frozen.

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What are some examples of properties of water?

gavmur [86]

Examples : cohesion , polarity , high specific heat , adhesion.<span />

4 0

3 years ago

How many moles of Na2SO4, are produced from 7.00 moles of (NH4)2SO4?

Harlamova29_29 [7]

Answer:

dumb stupid restarted illiterate STUPID students ARE JUST GOING TO DIE

8 0

2 years ago

A sample of water vapor has a volume of 3.15 L, a pressure of 2.40 atm, and a temperature of 325 K. What is the new temperature,

lara31 [8.8K]

Answer:

The answer to your question is: T2 = 235.44 °K

Explanation:

Data

V1 = 3.15 L V2 = 2.78 L

P1 = 2.40 atm P2 = 1.97 atm

T1 = 325°K T2 = ?

Formula

$\frac{P1V1}{T1} = \frac{P2V2}{T2}$

Process

T2 = (P2V2T1) / (P1V1)

T2 = (1.97x 2.78x 325) / (2.40 x 3.15)

T2 = 1779.895 / 7.56

T2 = 235.44 °K

4 0

2 years ago

Calculate the molality of isoborneol in the product if, a) the melting point of pure camphor is 179°C and the melting point take

tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

Melting point of pure camphor= T =179°C

Melting point of sample = $T_f$ = 165°C

Depression in freezing point = $\Delta T_f=?$

$\Delta T_f=T- T_f=179^oC-165^oC=14^oC$

Depression in freezing point is also given by formula:

$\Delta T_f=i\times K_f\times m$

$K_f$ = The freezing point depression constant

m = molality of the sample

i = van't Hoff factor

We have: $K_f$ = 40°C kg/mol

i = 1 ( organic compounds)

$\Delta T_f=14^oC$

$14^oC=1\times 40^oC kg/mol\times m$

$m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg$

The molality of isoborneol in camphor is 0.53 mol/kg.

8 0

2 years ago

I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0

2 years ago