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27 days ago

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NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te

mperature from 25.0 °C to 5.0 °C, how many grams of NH4NO3 should we use for every 100.0 g of water in the cold pack? Assume no heat was lost outside of cold pack, and the specific heat of the resulted solution was the same as water, or 4.184 J/(g•°C).

1 answer:

lorasvet [2.5K]27 days ago

6 0

Answer:

To lower the temperature of the solution from 25.0°C to 5.0°C, it is necessary to use 35.2g of NH₄NO₃ for every 100.0g of water.

Explanation:

In order to cool down the solution, we need:

4.184 J/g°C × (5.0°C - 25.0°C) × (100.0g + X) = -Y

8368 J + 83.68 J/gX = Y (1)

Here, x represents the grams of NH₄NO₃ required, and Y represents the energy needed to remove heat.

Furthermore, the energy Y becomes:

Y = 25700 J/mol × $\frac{1mol}{80,043g}$ X

Y = 321 J/g X (2)

Substituting (2) into (1)

8368 J + 83.68 J/g X = 321 J/g X

8363 J = 237.32 J/gX

X = 35.2g

This means 35.2g of NH₄NO₃ must be used for every 100.0g of water to achieve a temperature decrease from 25.0°C to 5.0°C.

I trust this information will be useful!

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Help on part "c": The forensic technician at a crime scene has just prepared a luminol stock solution by adding 19.0g of luminol

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1. The luminol stock solution has a molarity of 1.431 M.

2. In 2.00 L of the diluted spray, there are 0.12 moles of luminol.

3. The volume of the stock solution from Part A that contains the same number of moles present in the diluted solution from Part B is 83.86 ml.

Additional Information

Stoichiometry in Chemistry focuses on the quantitative aspects of chemical reactions, which includes calculations related to volume, mass, and the count of ions, molecules, and elements.

Key concepts in stoichiometry include:

1. Relative atomic mass
2. Relative molecular mass

This refers to the relative atomic mass of a molecule.

3. Mole

A mole represents the number of particles in a substance equivalent to the number of atoms in 12 grams of carbon-12.

1 mole = 6.02 × 10²³ particles.

The quantity of moles can also be derived by dividing mass (in grams) by either the relative mass of an element or the relative mass of a molecule.

$\large{\boxed{\bold{mol\:=\:\frac{grams}{ relative\:mass} }}}$

Luminol (C₈H₇N₃O₂) is utilized for detecting blood traces at crime scenes, due to its reaction with iron found in blood.

To prepare a luminol stock solution, 19.0 g of luminol is mixed into a total volume of 75.0 mL of water.

Thus, the molarity is calculated as:

1. Moles of Luminol

- the relative molecular mass of Luminol:

= 8.C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams/mol.

Thus, we have:

moles = grams / relative molecular mass.

$mole=\frac{19}{177}$

moles = 0.1073.

2. Molarity (M)

M = moles / volume

$M\:=\:{\frac{ 0.1703 }{75.10^{-3} L}$

M = 1.431.

b. The concentration of luminol in the spray bottle is 6.00 × 10⁻² M. Therefore, in a 2 L solution, the number of moles is:

moles = M × volume

moles = 6 × 10⁻² × 2

moles = 0.12.

c. The molarity of the stock solution (Part A) is 1.431 M.

The diluted solution (Part B) contains 0.12 moles of luminol.

To find the volume of the stock solution (Part A) that has the same moles as the diluted solution (Part B):

volume = moles / M

$volume\:=\:\frac{0.12}{1.431}$

volume = 0.08386 L = 83.86 mL.

Further Learning

moles of water you can generate

the amount of each atom in the chemical's formula

the proportion of hydrogen to oxygen atoms in 2 L of water

Keywords: mole, volume, molarity, Luminol, relative molecular mass

6 0

1 month ago

Read 2 more answers

An unknown element is found to have three naturally occurring isotopes with atomic masses of 35.9675 (0.337%), 37.9627 (0.063%)

Tems11 [2403]

Answer:

The correct choice for your inquiry is option A, Argon.

Explanation:

Isotope Atomic mass Percent (%)

1 35.9675 0.337

2 37.9627 0.063

3 39.9624 99.6

To calculate the average atomic mass: (Mass of isotope 1)(percent of 1) + (Mass of isotope 2)(percent of 2) + (Mass of isotope 3)(percent of 3)

Average atomic mass = (35.9675)(0.00337) + (37.9627)(0.00063) + (39.9624)(0.996)

Average atomic mass = 0.1212 + 0.0239 + 39.8025

Average atomic mass = 39.9476

Theoretical Atomic mass

a) Ar 39.95

b) K 39.10

c) Cl 35.45

d) Ca 40.08

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26 days ago

5. Gabi has plans with her friends to go to a concert on her birthday in 4 days. She is so excited that she wants to know how ma

Alekssandra [2719]

Consequently, she feels very anxious because she has 345600 seconds to wait. Explanation: 60 seconds make up 1 minute, and 60 minutes constitute an hour. Each hour has 3600 seconds (60*60) and 24 hours make up a day. Hence, 3600 seconds multiplied by 24 hours results in 1 day equating to 86400 seconds—therefore, over four days, we have 86400 * 4 equating to 345600.

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8 days ago

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

eduard [2520]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Starting temperature of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship 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

Next, determine the 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, the 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

Thus, the total entropy is the sum of the entropy changes of both blocks.

= -554.12 J/K + 647.49 J/K $\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

11 days ago

A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe

KiRa [2726]

Response:

The specific heat of the alloy $C_{a} = 0.37 \frac{KJ}{Kg K}$

Clarification:

Weight of the alloy $m_{a}$ = 25 gm

Initial temperature $T_{a}$ = 100°c = 373 K

Weight of the water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 25.32 °c = 298.32 K

Final temperature $T_{f}$ = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

$m_{a}$ $C_{a}$ [[ $T_{a}$ - $T_{f}$ ] = $m_{w}$ $C_w$ ( $T_{f}$ - $T_{w}$ )

25 × $C_{a}$ × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

$C_{a} = 0.37 \frac{KJ}{Kg K}$

This gives us the specific heat of the alloy.

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1 month ago