All categories

3 years ago

Does neon have a larger atomic radius than argon

Chemistry

1 answer:

Lera25 [3.4K]3 years ago

6 0

Answer:

Explanation:

Argon has a larger atomic radius

You might be interested in

What is sodium oxide when it’s broken down into its elements?

erik [133]

Answer:

Explanation:Typically, manufactured glass contains around 15% sodium oxide, 70% silica (silicon dioxide) and 9% lime (calcium oxide). The sodium carbonate "soda" serves as a flux to lower the temperature at which the silica mixture melts

8 0

3 years ago

List the three principal states of matter, give two examples of each

icang [17]

matter can take be one of the three states which are; (Solid, Liquid, Gas)

Think about water, it can be a Solid (Ice) a Liquid (not frozen obviously) and a Gas (when it evaporates)

7 0

3 years ago

Explain why metal powder is used instead of large piece of metal

Natasha2012 [34]

Answer:

The surface area increases, thus increasing the rate of reaction

Explanation:

When you use powder instead of solid metal, the surface area to volume ratio increases and so does the rate of reaction, because the chance of collisions between the particles increases as they can "touch" the metal more easily when it is looser, like powder.

8 0

3 years ago

How many miles are in 215 grams of water

Irina18 [472]

You're off to a good start, now find the mass of H2O and put it under I mol,

then multiply 1 mol over the mass of H2O by 215 grams

3 0

3 years ago

For parts a & b below, derive only the initial value problem set up.

Otrada [13]

Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

$\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )$

The rate of the sugar coming into the tank is 0

$\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min$

$So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100} \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C \\ \\ \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}$

Initial amount of sugar = 25 Pounds

Now; y(0) = 25

25 = Ce⁰

C = 25

So; y(t) = 25 $e^{-\dfrac{t}{100}$

Thus, the amount of sugar at any time t = $\mathbf{25 e^{^{-\dfrac{t}{100}}}}$

B) For tank B :

$\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\ \dfrac{dy}{50-y }= \dfrac{dt}{50}$

8 0

3 years ago