All categories

2 years ago

What amount of ice must be added to 250 g of water at 15°c to cool the water to 0°c and have no ice left

Chemistry

1 answer:

Juliette [100K]2 years ago

7 0

Answer:

$\mathbf{m= 47 \ g}$

Explanation:

The heat needed to melt ice $(q_1)$ + heat needed to cool water $(q_2)$ = 0

$m\Delta H_{fusion} + mc \Delta T = 0$

$m \times 333.55 J g^{-1} + 250 \ g \times 4.184 \ Jg^{-1 \ 0 } \times (-15.0 \ ^0 C) = 0$

$m \times 333.55 J g^{-1} =15690$

$m=\dfrac{15690 \ J }{333.55 J g^{-1} }$

$\mathbf{m= 47 \ g}$

You might be interested in

Calculate the specific heat of a substance when 63j of energy are transferred as heat to an 8.0 g sample to raise it temperature

Flura [38]

The formula for energy or enthalpy is:

E = m Cp (T2 – T1)

where E is energy = 63 J, m is mass = 8 g, Cp is the specific heat, T is temperature

63 J = 8 g * Cp * (340 K – 314 K)

6 0

3 years ago

Read 2 more answers

10) One mole of a diatomic ideal gas is initially at a temperature of 127 °C and

Vladimir79 [104]

Ok hold on one second

6 0

3 years ago

HELP ASAP 20 POINTS AAAA

pochemuha

Answer:

Me and my friends were going to do a science experiment. Jonny’s job was to make the HYPOTHESIS. He said the “ If we mix baking soda and vinegar together, the TEMPERATURE will go down.”

So then Molly mixed the baking soda and vinegar together and checked the TEMPERATURE. We all OBSERVED as the thermometer’s TEMPERATURE went down. “ your THEORY/ HYPOTHESIS was correct!” Exclaimed Molly.

Then the whole science GROUP let out with a cheer! And wrote the information down on their EXPERIMENTAL info chart. They took a microscope and looked at the mixture because they wanted to the the little PARTICLES in the mixture. Lily CONTROLED the microscope she zoomed in and out to see the particles.

Explanation:

i hope this helps:)

3 0

2 years ago

Hydrogen gas and fluorine gas will react to form hydrogen fluoride gas. What is the standard free energy change for this reactio

Marta_Voda [28]

Answer:

$\Delta G=-541.4kJ/mol$

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:

$H_2+F_2\rightarrow 2HF$

Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:

$\Delta G=2\Delta G_f^{HF}-(\Delta G_f^{H_2}+\Delta G_f^{F_2})\\\\\Delta G=2*-270.70kJ/mol-(0kJ/mol+0kJ/mol)\\\\\Delta G=-541.4kJ/mol$