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3 years ago

Va rog ! urgent! E pt chimie

Chemistry

1 answer:

n200080 [17]3 years ago

5 0

Answer:

An element

It stays shiny

Explanation:

Pure Gold is an element.

An element is a distinct substance that cannot be split-up into simpler substances. Such substances consists of only one kind of atom.

There are over a hundred elements known to date.

As an element, Gold is classified as a metal due to its very unique set of properties.

One of the indicator that gold does not react with oxygen is that it stays shiny. It does not give rusty look when exposed to air.

Substances that combines with oxygen have a rusty look or change appearance when expose to air. For example, iron.

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You have 0.5 L of air at 203 k in an expandable container at constant pressure. You heat the container to 273 k. What is the vol

Jobisdone [24]

You can use P1V1/T1 = P2V2/T2 but since pressure is constant is becomes V1/T1=V2/T2

V1=0.5 L
T1=203 K
T2=273 K
V2=unknown

0.5L/203 = V2/273
V2= 0.67 L so C

Hope this helps :)

4 0

3 years ago

Identify 2 ways to measure mass

Likurg_2 [28]

Answer:

The two ways to measure mass are subtraction and taring.

8 0

3 years ago

The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha

zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$25.9 = x*10.0$

$25.9 = 10.0\:x$

$10.0\:x = 25.9$

$x = \dfrac{25.9}{10.0}$

$\boxed{x = 2.59}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{53.3}{2^{2.59}}$

$m \approx \dfrac{53.3}{6.021}$

$\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

3 0

3 years ago

Which represents the correct equilibrium constant expression for the reaction below? Cu(s)+2Ag+(aq)<->Cu2+(aq)+2Ag(s)

balu736 [363]

K(eq) = concentration of products/concentration of reactant = [Cu+2] / [Ag+]^2
Activity of pure solid and liquid is taken as 1.
Hence last option is correct.
Hope this helps, have a great day ahead!
I hope this helps you alot
:)
:)
:0

4 0

3 years ago

Read 2 more answers

2KClO3 —> 2KCl2+ 3O2

garik1379 [7]

Answer:

False. The balanced equation should be

2KClO3-->2KCl + 3O2

it is a decomposition reaction.

3 0

3 years ago

Va rog ! urgent! E pt chimie ​

Va rog ! urgent! E pt chimie