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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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2 years ago

Michael Scott is driving through the parking lot and accidentally hits Meredith as she’s walking to her car. If his car makes co

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3 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,

Harlamova29_29 [7]

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

F = m a

a = F / m

We use kinematics to find lips

v = v₀ + a t

v = v₀ + (F / m) t

The moment is defined by

p = m v

The moment change

Δp = m v - m v₀

Let's replace the speeds in this equation

Δp = m (v₀ + F / m t) - m v₀

Δp = m v₀ + F t - m v₀

Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

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2 years ago