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liubo4ka
1 month ago
7

The density of a cube is 2.0gcm3. If the mass is 128 g, what is the length of each side of the cube?

Chemistry
2 answers:
eduard [2.7K]1 month ago
6 0

Answer:

4.0 cm

Explanation:

The formula for density defines it as a relation between mass and volume:

d = \frac{m}{V}

Assuming the side length of the cube is a, the volume can be represented as:

V = a^3

Inserting this into the density formula yields:

d = \frac{m}{a^3}

Rearranging for side a gives:

a = \sqrt[3]{\frac{m}{d}}

Given:

m = 128~g

d = 2.0~g/cm^3

We conclude with:

a = \sqrt[3]{\frac{128~g}{2.0~g/cm^3}} = 4.0~cm

eduard [2.7K]1 month ago
5 0

Density is calculated as Mass divided by Volume.

Volume can also be expressed as Mass divided by Density.

Volume becomes 2 divided by 128.

This results in Volume equal to 1/64.

Therefore, each side measures 4 cm.

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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
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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.

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Determine the percent yield for the reaction
Alekssandra [3086]

Answer:

The percent yield of Br₂ in this reaction amounts to 96.15%

Explanation:

The reaction's balanced stoichiometric equation is:

2 NaBr + 1 Cl₂ → 2 NaCl + 1 Br₂

To calculate the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

To determine the theoretical yield:

5.29 g of NaBr reacts with an excess of chlorine; therefore, NaBr is the limiting reagent, controlling the possible yield of products.

We convert 5.29 g of NaBr to moles.

Number of moles = (Mass)/(Molar mass)

Molar Mass of NaBr = 102.894 g/mol

Number of moles = (5.29/102.894) = 0.0514121329 = 0.05141 mole

According to the stoichiometry of the reaction:

2 moles of NaBr yield 1 mole of Br₂

Thus, 0.05141 mole of NaBr will produce (0.05141×1/2) mole of Br₂, which is 0.0257 mole of Br₂

Theoretical yield = Expected mass of Br₂ from the reaction

= (Number of moles) × (Molar mass)

Molar mass of Br₂ = 159.808 g/mol

Theoretical yield of Br₂ = 0.0257 × 159.808 = 4.108 g

Calculating the percent yield:

Percent yield = 100% × (Actual yield)/(Theoretical yield)

Actual yield = 3.95 g

Theoretical yield = 4.108 g

Percent yield = 100% × (3.95/4.108) = 96.15%

Hope this is helpful!!!

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I predict that there will be an increase in the seconds recorded in the time column. This is because, as more water is mixed with sodium thiosulfate, its concentration diminishes in each flask. Additionally, a lower concentration results in a slower reaction rate since fewer molecules of sodium thiosulfate means there are less frequent collisions with sulfuric acid. With fewer collisions occurring in the reaction, it takes a longer time for the reaction to complete, leading to increased time when sodium thiosulfate is diluted.

Explanation:

I can confirm that this explanation is accurate.

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