All categories

1 month ago

A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm3)

Chemistry

2 answers:

KiRa [2.9K]1 month ago

8 0

Answer:

The answer is 7160 cm

Explanation:

Given Data

Diameter = 1 mm

Length =?

Quantity of gold = 1 mol

Density = 17 g/cm³

Steps

1.- Obtain the atomic mass of gold

Atomic mass = 197 g

This means 197g ------------ 1 mol

2.- Find the volume of the wire

Density = mass/volume

Volume = mass/density

Volume = 197/17

Volume = 5.7 cm³

3.- Calculate the wire length

Volume = πr²h

Rearranging for h

h = volume / πr²

Radius = 0.05 cm

Substituting values

h = 5.7/(3.14 x 0.05²)

h = 5.7 / 0.0025

h = 7159.2 cm ≈ 7160 cm

eduard [2.7K]1 month ago

7 0

Answer:

The wire's length is 1477 cm

Explanation:

Step 1: Provided Data

Diameter = 1.00 mm

Gold moles = 1.00 mol

Gold density = 17.0 g/cm³

Molar mass of gold = 196.97 g/mol

Step 2: Determining volume

Volume = mass / density

Volume = 196.97 grams / 17.0 g/cm³

Volume = 11.6 cm³

Step 3: Finding length

Volume = π*r²*length

11.6 = π *(0.05)² *length

length = 1477 cm

<pThus, the wire's length is 1477 cm

You might be interested in

A specific amount of energy is emitted when excited electrons in an atom in a sample of an element return to the ground state.Th

castortr0y [3046]

The answer is C. The specific amount of energy released when excited electrons fall back to the ground state produces an emission spectrum. That energy is emitted as photons with precise wavelengths corresponding to the energy differences between levels. Because each element yields a characteristic set of wavelengths, the emission spectrum can be used to identify the element in the sample.

5 0

29 days ago

Read 2 more answers

66.667 mL of 3.000 M H2SO4 (aq) solution was neutralized by the stoichiometric amount of 4.000 M Al(OH)3 solution in a coffee cu

eduard [2782]

Answer:

$\large \boxed{\Delta_{\textbf{r}}H =\text{-4600 J$\cdot$ mol}^{-1}}$

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V: 66.667

c/mol·L⁻¹: 4.000 3.000

(b) Moles of H₂SO₄

$\rm \text{66.667 mL H$_{2}$}SO_{4} \times \dfrac{\text{3.000 mmol H$_{2}$SO}_{4}}{\text{1 mL H$_{2}$SO}_{4}} = \text{200.00 mmol H$_{2}$SO}_{4}$

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{2 mmol Al(OH)}_{3}}{\text{3 mmol H$_{2}$SO}_{4}}\\\\= \text{133.33 mmol Al(OH)}_{3}$

(d) Volume of Al(OH)₃

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{1 mL Al(OH)}_{3}}{\text{4 mmol H$_{2}$SO}_{4}} = \text{50.000 mL Al(OH)}_{3}$

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

Mass of sulfuric acid solution = 66.667 g

Mass of aluminium hydroxide solution = 50.000

TOTAL = 116.667 g

(b) ΔT

ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

$\begin{array}{ccccl}n\Delta H & +& mC \Delta T& = &0\\\text{0.066 667 mol }\times \Delta H& + & \text{116.667 g} \times 1.10 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 2.4 \, ^{\circ}\text{C} & = & 0\\0.066667 \Delta H \text{ mol} & + & \text{310 J} & = & 0\\&&0.066667 \Delta H \text{ mol} & = & \text{-310 J} & & \\\end{array}\\$

$\begin{array}{ccccl}& &\Delta H & = & \dfrac{\text{-310 J}}{\text{0.066667 mol}}\\\\& &\Delta H & = & \textbf{-4600 kJ/mol}\\\end{array}\\\large \boxed{\mathbf{\Delta_{\textbf{r}}H} =\textbf{-4600 J$\cdot$ mol}^{\mathbf{-1}}}$

This result appears nonsensical, but it is derived from your given figures.

6 0

1 month ago

Thallium-207 decays exponentially with a half life of 4.5 minutes. if the initial amount of the isotope was 28 grams, how many g

Anarel [2989]

An exponential decay law is generally expressed as: A = Ao * e ^ (-kt) => A/Ao = e^(-kt) Half-life time => A/Ao = 1/2, and t = 4.5 min => 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154. Now substituting k, Ao = 28g, and t = 7 min to determine the remaining grams of Thallium-207 gives: A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g. Final answer is 9.5 g.

7 0

1 month ago

1. Bailey wants to find out which frozen solid melts the fastest: soda, gatorade, or orange juice. She pours each of the three l

VMariaS [2998]

First scenario:

IV: soda, gatorade, orange juice, and water

DV: state of the liquids listed above

Control: freezer and ice tray

Second scenario:

IV: laundry detergent, water

DV: cleanliness of the squares post-wash

Control: chocolate, cloth type, cloth squares