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3 years ago

A 10.0-gram sample of liquid water at 23.0°C absorbs 209 joules of heat.

Chemistry

1 answer:

Aliun [14]3 years ago

8 0

Answer:

The final temperature of the water is 28.0 °C

Explanation:

Step 1: Data given

Mass of liquid water = 10.0 grams

Temperature = 23.0 °C

Heat absorbed = 209 Joules

Since heat was absorbed by the water, you must have a positive value for

Δ T

Step 2: Calculate final temperature

q = m*c* ΔT

⇒ with m = the mass of the water = 10.0 grams

⇒ with c = the specific heat of water = 4.184 J/g°C

⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 23.0 °C

⇒ with q = the heat absorbed = 209 Joule

209 = 10.0 * 4.184 * ΔT

ΔT = 5

ΔT = 5 = T2 - 23

T2 = 28 °C

The final temperature of the water is 28.0 °C

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Answer:

47.01 g/mol is molar mass

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4 years ago

Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W

nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

H2= (1-x)Haq+XHvap.........1

Putting the values in 1

2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

x = 1904.976 (J/g)/1998.576 (J/g)

x = 0.953

So, the quality of the wet steam is 0.953

7 0

3 years ago

Read 2 more answers

Write the complete atomic structure for the Carbon atom. The atomic and mass numbers for carbon are 6 and 12 reapectively.

sergiy2304 [10]

Answer:

C-12 or C with a 12 superscripted on Upper left and 6 Subscripted on bottom left

Explanation:

Isotopic notation

4 0

3 years ago

a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a bu

Alex17521 [72]

Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

3 0

3 years ago

Can someone please help????nThis is sooooo hard

Rashid [163]

Answer:

See explanation.

Explanation:

I highly suggest you watch OChem Tutor's videos on IUPAC nomenclature because the actual naming would take a lot of time to teach in text-based format. But here is how to name them:

1) I think there are two seperate pictures for number 1. The molecule on the left is 1-pentene and the one on the right is 4-methyl-1-pentene. If the whole thing is one molecule but there is just a bond missing where the red marker numbers are, that molecule would be 9-methyl-1,6-decadiene.

2) 4-methyl-2-pentene

3) 2,4-octadiene

4) 1,5-nonadiene

5) 2,5-dimethyl-3-hexene

6) 3,6-dimethyl-2,4-heptadiene

7) 2,5,5-trimethyl-2-hexene

6 0

2 years ago