Help me with 4,5 plz

Help please ill give brainliest (mind the "Because" i was just writing something

Citrus2011 [14]

Answer:

never start off with because

Explanation:

thermal energy is added to the cube for it to melt since it wasnt in nothing cold even thought is would only stay ice n the freezer not the refrigrator

5 0

3 years ago

Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

ira [324]

Explanation:

The given data is as follows.

m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)

Initial temp. of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Initial temp. of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

So, heat released by block 1 = heat gained by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

Convert temperature into kelvin as (50 + 273) K = 323 K.

Also, we know that the relation between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Now, calculate entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Hence, total entropy will be sum of entropy change of both the blocks.

$\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= -554.12 J/K + 647.49 J/K

= 93.37 J/K

Thus, we can conclude that for the given reaction $\Delta H$ is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

3 years ago

What are three limitations of current cloaking technology.

kirill [66]

Explanation:

1.undetectable to electromagnetic waves

2.hiding an object from an illumination containing diffre t wave lengths become difficult as the object sizes grow.

3. reduce the scattering by two orders.

4 0

3 years ago

The strongest light left over from the big bang is called background radiation and was given off as which type of wave?

Sonja [21]

It would be Microwaves. The cosmic microwave background (CMB, CMBR), in Big Bang cosmology, is electromagnetic radiation as a remnant from an early stage of the universe, also known as "relic radiation".

8 0

3 years ago

The mass percentage of hydrochloric acid within a solution is 28.00%28.00% . Given that the density of this solution is 1.1411.1

sashaice [31]

Answer:

8.76M

Explanation:

Given that

Mass from the density = 1141g

According to the given situation the computation of molarity of the solution is shown below:-

we will took HCL solution which is 1000mL

HCl = 28% by mass

So,

Mass of HCl in 1-litre solution is

$= \frac{28}{100} \times 1141$

Which gives the result of molar mass HCI is

= 319.48g /mol

Now,

Molarity is

$= \frac{319.48}{36.45}$

Which gives results of molarity is

= 8.76M

8 0

3 years ago