Reaction rate lab report

Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a

lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT

Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g

Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT

Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g

Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) THIS IS THE RIGHT ANSWER (the more negative= the more released)

7 0

3 years ago

You are carrying out an experiment in the lab to study the glycolysis pathway. To do this, a liver extract (which is capable of

Virty [35]

Answer:

When C1 is labeled in glucose, it ends up in the methyl group of pyruvate.

Aldolase cleaves a hexose into two trioses.

[See the image attached].

Asterisk indicates the label.

When C1 is labeled in glucose, it ends up in the carboxyl group of pyruvate.

5 0

3 years ago

.....................?

liraira [26]

Answer:

Look at the periodic table of elements

Explanation:

Group 1 is the most reactive and to the right

Group 18 is least reactive and all the way to the left. The rest are in between. The groups go vertical.

You'd be able to solve with this information youre welcome

4 0

2 years ago

How many atoms are in 5.70 x 10^32 mol of Rn?

Alex73 [517]

To find amount of atoms from mol, multiply the mole amount by Avogadro’s number

5.70x10^32 x 6.02x10^23

= 3.43x10^56 atoms

5 0

2 years ago

Calculate the radius of tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm, and an atomic wei

Ivahew [28]

Answer:

The radius of tantalum (Ta) atom is $R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

Explanation:

From the Body-centered cubic (BBC) crystal structure we know that a unit cell length a and atomic radius R are related through

$a=\frac{4R}{\sqrt{3} }$

So the volume of the unit cell $V_{c}$ is

$V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}$

We can compute the theoretical density ρ through the following relationship

$\rho=\frac{nA}{V_{c}N_{a}}$

where

n = number of atoms associated with each unit cell

A = atomic weight

$V_{c}$ = volume of the unit cell

$N_{a}$ = Avogadro’s number ( $6.023 \times 10^{23}$ atoms/mol)

From the information given:

A = 180.9 g/mol

ρ = 16.6 g/cm^3

Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.

We can use the theoretical density ρ to find the radio R as follows:

$\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}$

Solving for R

$\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}$

Substitution for the various parameters into above equation yields

$R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

7 0

3 years ago