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3 years ago

The mass of a single gold atom is 3.27×10-22 grams. how many gold atoms would there be in 236 milligrams of gold?

2 answers:

8 0

The amount of gold atoms could be calculated by dividing the total weight of the gold with the mass of a single gold atom. Just convert the given weight to grams then divide it with 3.27x10^-22 grams. The answer would be 7.22x10^20.

Arisa [49]3 years ago

3 0

Answer:

6.38*10²⁰ atoms of gold

Explanation:

236mg of gold = 236*10^-3g = 0.236g.

Now from the question,

1 atoms contains = 3.7*10^-22g.

X atoms will contain = 0.236g of gold.

Now we'll cross multiply and make x the subject of formula to find the amount of gold in 0.236g of gold.

X = (0.236 * 1) / 3.7*10^-22

X = 6.378*10^20 atoms

0.236g of gold contains 6.378*10^20 atoms of gold

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An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag

Afina-wow [57]

Answer:

The induced emf in the loop is $7.35\times 10^{-4}\ V$

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, $A=a^2=(0.305)^2=0.0930\ m^2$

Circumference of the loop,

$C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m$

Area of circle,

$A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2$

The induced emf is given by :

$\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V$

So, the induced emf in the loop is $7.35\times 10^{-4}\ V$

8 0

3 years ago

You are helping two friends from our class with a physics problem where a cart is pushed up a ramp. In examining the motion of t

stiks02 [169]

Answer: Acceleration will have 2 components, vertical and horizontal.

Net-vertical component can be positive, zero or negative depending upon the magnitude of the upward component of the applied acceleration.

Net-horizontal acceleration will be equal to the horizontal component of the applied acceleration.

Explanation:

Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.

<u>For vertical acceleration:</u>

If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.

In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.

<u>For horizontal acceleration:</u>

This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.

But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.