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Leto [7]
3 years ago
13

(I will give a brainliest) State the relationship between phase changes and temperature

Chemistry
1 answer:
Alex73 [517]3 years ago
8 0

Answer:

The relationship is the "<em>variation</em><em> </em><em>of</em><em> </em><em>vapour</em><em> </em><em>pressure</em><em> </em><em>with</em><em> </em><em>temperature</em><em>"</em><em> </em>or <em>"</em><em> </em><em>the</em><em> </em><em>effect</em><em> </em><em>of</em><em> </em><em>vapour</em><em> </em><em>pressure</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>melting</em><em> </em><em>and</em><em> </em><em>boiling</em><em> </em><em>points</em><em> </em><em>in</em><em> </em><em>a</em><em> </em><em>phase</em><em>"</em>

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QUESTION 17
borishaifa [10]

Question 17:

False; cutting hair would change what it looks like but, braiding it keeps it the same substance it is.


Question 18 :

My best guess would either be A or D. I would lean more towards D because if there are different mixtures then it depends on what you are mixing.


Hope it helps in some way. <3

5 0
3 years ago
Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHc
jeka94

Answer:

T_f = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

T_i = 21.5°C

C_{calorimeter} = 6.15 kJ/°C

T_f = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate ×  \frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

=  0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter =  C_{calorimeter} (T_f-T_i} )

21.84 kJ =6.15 kJ/°C * (T_f-21,5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - 132.225 kJ

21.84 KJ + 132.225 kJ = (6.15 kJ/^0C * T_f)

154.065 kJ = (6.15 kJ/^0C * T_f)

T_f = \frac{154.065kJ}{6.15kJ/^0C}

T_f =25.05°C

4 0
4 years ago
Write at least three characteristics of acids and three characteristics of bases. These characteristics should be what makes the
ki77a [65]
Characteristics of acid

-It tastes sour
-It reacts with metals and carbonates
-It turns blue litmus paper red

Characteristics of base

-It tastes bitter
-It feels slippery
-It turns red litmus paper blue.
3 0
3 years ago
A flask with a volume of 250.0 mL contains air with a density of 1.164 g/L. What is the mass of the air contained in the flask?
satela [25.4K]

Answer:

convert 250.0 mL in Liters :250. 0 / 1000 = 0,25 LDensity = 1.240 g/LMass

Explanation:

6 0
2 years ago
CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the
nirvana33 [79]

Answer:

The answer to your question is   P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L                        Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C               Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa                    Pressure 2 = P2 =  ?

Process

- To solve this problem use the Combined gas law.

                     P1V1/T1 = P2V2/T2

-Solve for P2

                     P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

                       1000 ml -------------------- 1 l

                         855 ml --------------------  x

                         x = (855 x 1) / 1000

                         x = 0.855 l

-Substitution

                    P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

                    P2 = 377600 / 141.075

-Result

                   P2 = 2676.6 kPa

3 0
3 years ago
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