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2 years ago

How many gold atoms are in a 1kg gold bar?

Physics

2 answers:

Naya [18.7K]2 years ago

7 0

Answer:

The answer is a 1 kilo gold bar has 32.15 ounces of gold or 1000 grams.

Explanation:

SpyIntel [72]2 years ago

3 0

Answer:

. You use the fact that 1 mole of any substance contains exactly 6.022⋅1023 atoms or molecules of that substance - this is known as Avogadro's number. In your case, 1 mole of gold will have exactly 6.022⋅1023 atoms of gold

Explanation:

Well, what is the molar mass of gold metal?

My (and your!) Periodic Table tells me that it has a molar mass of 196.97∙g∙mol−1.

And so (i) we determine the molar quantity of gold, i.e.

Moles of gold≡1000∙g196.97∙g∙mol−1≡5.077∙mol.

And (ii), since we KNOW that 1∙mol represents 6.02214076×1023 individual items…we take the product….

5.077∙mol×6.02214076×1023∙mol−1≡3.057×1024∙individual gold atoms.

Given that gold is worth $48–30 USD∙g−1 how much is this mass worth? And what is the price for an ATOM of gold?

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A girl runs once around a circular track with a radius of 100m at speed of 10m/s.

goblinko [34]

Answer:

0 m

Explanation:

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5 0

2 years ago

PLEASE HELP!! At what temperature will silver have a resistivity that is four times the resistivity of tungsten at room temperat

Svetach [21]

Consider 20 deg.C. as room temperature.

From tables,
Silver has a resistivity of 1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m

At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.

The resistivity of silver will be 4 times that of tungsten (at room temperature) when
1.6*10^-8(1 + 0.0038T) = 4*5.6*10^-8
1 + 0.0038T = 14
T = 13/.0038 = 3421 deg.K approx

Answer: 20 + 3421 = 3441 °C

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3 years ago

During a plane showcase, a pilot makes circular "looping" with a speed equal to the sound speed (340 m/s). However, the pilot ca

Dmitriy789 [7]

Answer:

1472.98 m

Explanation:

Data provided:

Speed of circular looping, v = 340 m/s

Acceleration, a = 8g

here,

g is the acceleration due to the gravity = 9.81 m/s²

Now,

the centripetal acceleration is given as,

$a=\frac{v^2}{r}$

r is the radius of the loop

on substituting the respective values, we get

$8\times9.81=\frac{340^2}{r}$

r = 1472.98 m

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10 months ago