Ask question

Ask question

All categories

6 days ago

10

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

rature of the water rose to a final value of 27.18 oC. Neglecting heat losses to the room and the heat capacity of the beaker itself, what is the specific heat of the alloy

1 answer:

KiRa [971]6 days ago

4 0

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.

You might be interested in

An atom of oxygen has an atomic number of 8 and a mass number of 18. How many of each type of subatomic particle does it contain

VMariaS [1037]

Answer:

8 protons, 8 electrons, and 10 neutrons

Explanation:

6 0

10 days ago

Read 2 more answers

A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 gra

lorasvet [956]

Answer:

C₂H₅O₂

Explanation:

From the information provided in the question, we have the following details:

Mass of compound = 0.25 g

Mass of CO₂ = 0.3664 g

Mass of H₂O = 0.15 g

Empirical formula =?

Now, let’s calculate the masses of carbon, hydrogen, and oxygen within the compound as follows:

For Carbon (C):

Mass of CO₂ = 0.3664 g

Molar mass of CO₂ = 12 + (2×16) = 44 g/mol

Mass of C = 12/44 × 0.3664

Mass of C = 0.1

For Hydrogen (H):

Mass of H₂O = 0.15 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H = 2/18 × 0.15

Mass of H = 0.02 g

For Oxygen (O):

Mass of C = 0.1 g

Mass of H = 0.02 g

Mass of compound = 0.25 g

Mass of O =?

Mass of O = (Mass of compound) – (Mass of C + Mass of H)

Mass of O = 0.25 – (0.1 + 0.02)

Mass of O = 0.25 –0.12

Mass of O = 0.13 g

In conclusion, we will now find the empirical formula for the compound:

C = 0.1

H = 0.02

O = 0.13

Next, we divide by their respective molar mass:

C = 0.1 / 12 = 0.0083

H = 0.02 / 1 = 0.02

O = 0.13 / 16 = 0.0081

Then we divide by the smallest value:

C = 0.0083 / 0.0081 = 1

H = 0.02 / 0.0081 = 2.47

O = 0.008 / 0.008 = 1

Finally, we multiply by 2 to present in whole numbers:

C = 1 × 2 = 2

H = 2.47 × 2 = 5

O = 1 × 2 = 2

Therefore, the empirical formula for the compound is C₂H₅O₂

5 0

8 days ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

lions [985]

Answer: The empirical formula and molecular formula for the analyzed organic compound are $C_9H_{12}O$ and $C_{18}H_{24}O_2$

Explanation:

The combustion chemical equation for a hydrocarbon containing carbon, hydrogen, and oxygen is:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where 'x', 'y', and 'z' represent the subscripts for Carbon, Hydrogen, and Oxygen.

We have the following data:

Mass of $CO_2=39.01g$

Mass of $H_2O=10.65g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For carbon mass calculation:

12 grams of carbon are found in 44 grams of carbon dioxide.

Thus, in 39.01 grams of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ grams of carbon will be present.

For hydrogen mass calculation:

In 18 grams of water, 2 grams of hydrogen are contained.

Therefore, in 10.65 grams of water, $\frac{2}{18}\times 10.65=1.18g$ grams of hydrogen will be present.

The oxygen mass in the compound = (13.42) - (10.64 + 1.18) = 1.6 grams.

To derive the empirical formula, you need to perform a few steps:

Step 1: Convert the provided masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

Step 2: Determine the mole ratio of the elements.

Each mole value is divided by the smallest mole value, which is 0.1 moles, to find the mole ratio.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

Step 3: Use the mole ratio values as subscripts.

The ratio of C: H: O = 9: 12: 1

The empirical formula for the compound is $C_9H_{12}O$

To find the molecular formula, it’s necessary to ascertain the valency, which is then multiplied by each elemental subscript of the empirical formula.

The formula used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

Given:

Mass of the molecular formula = 272.38 g/mol

Mass of the empirical formula = 136 g/mol

Substituting values into the equation gives:

$n=\frac{272.38g/mol}{136g/mol}=2$

Multiplying the determined valency with the empirical formula’s element subscripts results in:

$C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2$

Therefore, the forms of the organic compound are $C_9H_{12}O$ and $C_{18}H_{24}O_2$

6 0

2 days ago

An experiment in chm 2045 requires students to prepare a 1.0 M aqueous solution of potassium phosphate.Jennifer fills a 1.0 lite

alisha [964]

Answer:

Joe correctly mixed the solution.

Explanation:

When evaluating both procedures, it's evident that both Jennifer and Joe weighed the same amount of potassium phosphate, which isn’t the variable here.

The difference is that Jennifer added the solid to 1.0 liters of water, resulting in a final volume greater than 1.0 L, thus her concentration will be lower than 1.0 M.

Joe's solution has a final volume of 1.0 L, which is why his preparation is accurate.

6 0

2 days ago

A Gas OCCUPIES 525ML AT A PRESSURE OF 85.0 kPa WHAT WOULD THE VOLUME OF THE GAS BE AT THE PRESSURE OF 65.0 kPa

castortr0y [923]

Boyle's law pertaining to ideal gases states that the volume of a gas relates inversely to its pressure when temperature is constant. According to this principle, the relationship between pressure and volume can be expressed as:

$PV=Constant$

This implies:

$P_{1}V_{1}=P_{2}V_{2}$

Utilizing this equation enables us to determine the volume of gas given a specific pressure:

P₁=Initial pressure

V₁=Initial volume

P₂=Final pressure

V₂= Final volume

In this case, the initial pressure P₁ is noted as 85.0 kPa.

The initial volume V₁ is provided as 525 mL.

The final pressure P₂ is recorded as 65.0 kPa.

$P_{1}V_{1}=P_{2}V_{2}$

Thus,

V_{2}=85×525÷65

=686 mL

Consequently, the gas volume will amount to 686 mL.

8 0

7 days ago

Read 2 more answers