In ΔDEF, the measure of ∠F=90°, FE = 16, DF = 63, and ED = 65. What ratio represents the tangent of ∠E?

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

2 years ago

Paintball is a popular recreational activity that uses a metal tank of compressed carbon

Dovator [93]

Answer:

Explanation:

Here we have the mass of CO₂ added = 340 g

From

$Number \, of \, moles = \frac{Mass}{Molar \, mass}$

We have, where the molar mass of CO₂ is 44.01 g/mol

Therefore,

$Number \, of \, moles = \frac{340}{44.01} = 7.73 \, \, \, moles$

71. Included drawing attached

72. Here we have the pressure of the gas given by Charles law which can be resented as follows;

$\frac{P_1}{P_2} =\frac{T_1}{T_2}$

Where:

P₁ = Initial pressure = 6.1 atmospheres

P₂ = Final pressure

T₁ = Initial Temperature = 293 K

T₂ = Initial Temperature = 313 K

Therefore,

$P_2= T_2 \times \frac{P_1}{T_1} = 313 \times \frac{6.1}{293} = 3.312 \, \, \, atmospheres$

5 0

2 years ago

As Danny was pouring cereal for his breakfast, he noticed that the cereal box says that the cereal contains 5 milligrams of iron

Vinil7 [7]

Answer:

Correct, because B it is reported to the nearest miligram

Explanation:

4.6 rounded up is 5

6 0

2 years ago

Stephanie has been doing research on how petroleum is formed. She says that oxygen must be present while the tiny plants and ani

madreJ [45]

Answer:

No, Stephanie is incorrect. Formation of petroleum cannot take place under the presence of oxygen.

Explanation:

Since, the petroleum is fossil product. Fossil fuel are formed under high pressure and temperature with absence of oxygen for longer period. so the way she is performing is completely incorrect. With the presence of oxygen in no way petroleum will be formed. The temperature and pressure should be in different combination for the formation of the petroleum. Along with the layers of sediments to maintain the pressure is required.

5 0

2 years ago

Read 2 more answers

The chemical formula for artificial sweetener is C7H5NO3S. Match the elements with the numbers found in one molecule of artifici

klio [65]

its is 1.hydrogen 2. oxygen 3. argon 4. carbon

7 0

2 years ago