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4 years ago

Does mint actually cool things down

1 answer:

3 0

The cooling of a mint is just a sensation if you were to put mints into warm water the water would remain warm because it is just a cooling sensation. I hope this helps:)

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In the 1 H NMR spectrum of chloroethane the methylene group is split into a quartet by the α and β nuclear spins of the protons

erastova [34]

Answer:+1/2 and -1/2.

Explanation:the presence of an external magnetic field (B0), two spin states exist, +1/2 and -1/2.

The magnetic moment of the lower energy +1/2 state is aligned with the external field, but that of the higher energy -1/2 spin state is opposed to the external field.

7 0

3 years ago

HELP QUICK!!

worty [1.4K]

I believe it’s A, they form whole-number ratios in the compound

4 0

4 years ago

Read 2 more answers

A lab scientist cools a liquid sample of water (2.6 kg) at 0.00°C to -192°C. The water turns to ice as this temperature change o

soldier1979 [14.2K]

Answer: The heat released for the given process is -1892 kJ

Explanation:

The processes involved in the given problem are:

$1.)H_2O(l)(0^oC,273K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(s)(-192^oC,81K)$

Pressure is taken as constant.

To calculate the amount of heat released at same temperature, we use the equation:

$q=m\times L_{f,v}$ ......(1)

where,

q = amount of heat released = ?

m = mass of water/ice

$L_{f,v}$ = latent heat of fusion or vaporization

To calculate the amount of heat released at different temperature, we use the equation:

$q=m\times C_{p,m}\times (T_{2}-T_{1})$ .......(1)

where,

q = amount of heat released = ?

$C_{p,m}$ = specific heat capacity of medium

m = mass of water/ice

$T_2$ = final temperature

$T_1$ = initial temperature

Calculating the heat absorbed for each process:

For process 1:

Converting the latent heat of fusion in J/kg, we use the conversion factor:

1 kJ = 1000 J

So, $(\frac{-334kJ}{1kg})\times (\frac{1000J}{1kJ})=-334\times 10^3J/kg$

We are given:

$m=2.6kg\\L_f=-334\times 10^3J/kg$

Putting values in equation 1, we get:

$q_1=2.6kg\times (-334\times 10^3J/kg)=-868400J$

For process 2:

We are given:

$m=2.6kg\\C_{p,s}=2050J/kg.K\\T_1=273K\T_2=81K$

Putting values in equation 2, we get:

$q_2=2.6kg\times 2050J/kg.K\times (81-(273))^oC\\\\q_2=1023360J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[-868400+(-1023360)]J=-1891760J=-1891.76kJ\approx -1892kJ$

Hence, the heat released for the given process is -1892 kJ

3 0

3 years ago

A few substances are arranged in the increasing order of ‘forces of attraction’ between their particles. Which one of the follow

gavmur [86]

Answer:

Explanation:

PO WATER,AIR,WIND

7 0

3 years ago

Please choose one of the choices

Georgia [21]

Answer:

A. the law of constant composition

Explanation:

The molecules in the container would have the same composition because they would have traded around atoms until an equilibrium was reached with every molecule having 1 Hydrogen and 1 Chlorine.

7 0

3 years ago