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14 days ago

Calculate the ratio of the velocity of helium atoms to the velocity of neon atoms at the same temperature.

Chemistry

1 answer:

Tems11 [2.4K]14 days ago

8 0

Answer:

vHe / vNe = 2.24

Explanation:

To determine the velocity of an ideal gas, one should apply the formula:

v = √3RT / √M

In this equation, R represents the gas constant (8.314 kgm²/s²molK); T refers to temperature, and M indicates the molar mass of the gas (4x10⁻³kg/mol for helium and 20.18x10⁻³ kg/mol for neon). Hence:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio simplifies to:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

vHe / vNe = 2.24

I hope it assists you!

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Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter

KiRa [2726]

Answer:

thickness is 0.29 cm

Explanation:

To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.

Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).

We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.

The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.

This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.

Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.

So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.

Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].

Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].

30.1635 * 3/(4 * 3.14) = 27 - r^3.

Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.

Therefore, r = 2.7051 cm.

This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.

Rounding to two significant figures yields

the thickness = 0.29 cm.

8 0

1 month ago

In living organisms, C-14 atoms disintegrate at a rate of 15.3 atoms per minute per gram of carbon. A charcoal sample from an ar

lions [2653]

Answer:

The result is "4,241.17 years"

Explanation:

The disintegration rate for C-14 atoms is indicated in $15.3 \frac{atoms}{min-g}$

The dissolution rate of the sample is given by $9.16 \frac{atoms} {min-gram}$

The C-14 proportion within the sample can be determined as per $= \frac {9.16}{15.3} \\\\ = 0.5987$

With a half-life of 5730 years.

Now, we need to compute the number of half-lives (n) that are applicable:

$(\frac{1}{2})^n= A\\\\A= fraction of C-14, which is remaining \\\\(\frac{1}{2})^n= 0.5987 \\\\ n \log 2 = - \log 0.5987\\\\$

$\therefore \\\\ \Rightarrow n= \frac{0.227}{0.3010} \\\\ = 0.740\\$

Thus, the age of the sample is represented as = $n \times\ half-life$

$= 0.740 \times 5730 \ years \\\\=4241.17 \ years\\\\$

6 0

1 month ago

A hardware salesman measures the mass of a box containing

lorasvet [2542]

Step $1$ : To find the mass of a single washer, divide the total measured mass by the number of washers $\frac{0.4168}{1000} =0.0004168\ Kg$ $0.0004168kg=4.168*10^{-4} kg$ . Step $2$ : Convert kilograms to milligrams knowing $1 kg=1*10^{6} mg$ ; thus $4.168*10^{-4}*10^{6} =4.168*10^{2}mg=416.8 mg$ resulting in the final answer $416.8 mg$ .

8 0

1 month ago

Read 2 more answers

(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

KiRa [2726]

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

$Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}$

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

7 0

1 month ago

If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

Anarel [2605]

Answer:

The amount of calcium sulfate that precipitates is 6.14 grams.

Explanation:

Step 1: Provided data

We are mixing 500.0 mL of 0.10 M Ca^2+ with 500.0 mL of 0.10 M SO4^2−

The Ksp for CaSO4 is 2.40×10^−5.

Step 2: Determine moles of Ca^2+

Moles of Ca^2+ = Molarity of Ca^2+ * Volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles of Ca^2+ = 0.05 moles

Step 3: Determine moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Compute total volume

Total volume = 500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Compute Q

Q = [Ca2+] [SO42-]

[Ca2+] = 0.050 M and [SO42-]

Qsp = (0.050)(0.050) = 0.0025 >> Ksp

This indicates that precipitation will take place.

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M (molar solubility)

Step 7: Determine total dissolved CaSO4

Total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 g/mol = 0.667 g

Step 8: Calculate initial mass of CaSO4

Initial moles of CaSO4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate precipitate mass

6.807 - 0.667 = 6.14 grams.

The mass of calcium sulfate that will emerge as a precipitate is 6.14 grams.

5 0

27 days ago