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Alex73 [517]
2 years ago
11

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:
allochka39001 [22]2 years ago
8 0

Answer:

The answer to your question is 7160 cm

Explanation:

Data

diameter = 1 mm

length = ?

amount of gold = 1 mol

density = 17 g/cm³

Process

1.- Get the atomic mass of gold

Atomic mass = 197 g

then, 197g ------------ 1 mol

2.- Calculate the volume of this wire

density = mass/volume

volume = mass/density

volume = 197/17

volume = 5.7 cm³

3.- Calculate the length of the wire

Volume = πr²h

solve for h

h = volume /πr²

radius = 0.05 cm

substitution

h = 5.7/(3.14 x 0.05²)

h = 5.7 / 0.0025

h = 7159.2 cm ≈ 7160 cm

just olya [345]2 years ago
7 0

Answer:

The length of this wire is 1477 cm

Explanation:

Step 1: Data given

diameter = 1.00 mm

Number of moles of gold = 1.00 mol

Density of gold = 17.0 g/cm³

Molar mass of gold = 196.97 g/mol

Step 2: Calculate volume

Volume = mass / density

Volume = 196.97 grams /17.0 g/cm³

Volume = 11.6 cm³

Step 3: Calculate length

Volume =  π*r²*length

11.6 =  π *(0.05)²  *length

length = 1477 cm

The length of this wire is 1477 cm

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lana66690 [7]

Answer:

Wavelength of this beam of light: \rm 4.39\times 10^{-7}\; m.

Explanation:

The speed of light in vacuum is approximately \rm 2.998\times 10^{8}\;m \cdot s^{-1}.

Light behaves like a wave. The wavelength of a wave is equal to the distance that it travels (in the given medium) in each period of oscillation.

On the other hand, the frequency of a wave is the number of periods in unit time. 1\rm \; Hz means one oscillation per second. The frequency of this particular wave is \rm 6.83\times 10^{14}\; Hz. In other words, there are 6.83\times 10^{14} oscillations in each second.

The period of oscillation will be equal to

\displaystyle t = \frac{1}{f} = \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}}.

In that period of time, a beam of light in vacuum would have traveled  

\displaystyle \rm 2.998\times 10^{8}\; m\cdot s^{-1} \times \frac{1}{\rm 6.83\times 10^{14}\; s^{-1}} = 4.39\times 10^{-7}\; m.

In other words, if this beam of light of frequency \rm 6.83\times 10^{14}\; Hz is in vacuum, its wavelength will be equal to \rm 4.39\times 10^{-7}\; m.

8 0
2 years ago
We have an object with a density of 620 g/ cm3 and a volume of 75 cm3. What is the mass of this object?
Anna11 [10]
M=D*V
D=620 g/cm³
V=75 cm³

m= 620 g/cm³ * 75 cm³=46500 g
m=46500g
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2 years ago
When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

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