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Ket [755]
3 years ago
15

In right triangle EFG, mE = 25. In right triangle HJK, mH = 25. Which similarity postulate or theorem proves that EFG and HJK ar

e similar?
Chemistry
2 answers:
Rus_ich [418]3 years ago
6 0
The angle- angle similarity postulate. It states that if a triangle has two equal corresponding angles the angles are similar.
bagirrra123 [75]3 years ago
4 0

The postulate or theorem which proves that EFG and HJK triangles are similar is Angle-Angle  (AA)

according to it

In case of two triangles if two pairs of corresponding angles are congruent, the triangles are considered to be similar triangles.

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When two forces acting on an object are equal, they are
TEA [102]
The answer is balanced forces
7 0
2 years ago
A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

Learn more:

  • brainly.com/question/25475410
  • brainly.com/question/12625048
6 0
2 years ago
a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp
Snowcat [4.5K]

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH =  11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity =  4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

Learn more about heat of solution:brainly.com/question/24243878

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4 0
1 year ago
Explain why the process of dissolution can be exothermic or endothermic (heat of solution) with different solutes
RUDIKE [14]

Answer:

The process of dissolving is exothermic when more energy is released when water molecules “bond” to the solute than is used to pull the solute apart. Because more energy is released than is used, the molecules of the solution move faster, making the temperature increase.

Project the image Endothermic Dissolving.

The process of dissolving is endothermic when less energy is released when water molecules “bond” to the solute than is used to pull the solute apart. Because less energy is released than is used, the molecules of the solution move more slowly, making the temperature decrease.

4 0
3 years ago
Calculate the heat energy released when 21.1 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point.
PSYCHO15rus [73]
Heat capacity yea urjjrjrjrbdnsnnend
3 0
2 years ago
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