How many moles of copper are in 6,000,000 atoms of copper?

Chemistry

1 answer:

crimeas [40]3 years ago

7 0

Answer: There will be 9.9632 × 10⁻¹⁸ moles of Copper in 6,000,000 atoms of Copper.

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As the particles of a solid undergo sublimation , they?

vladimir2022 [97]

They are particular solids.

4 0

3 years ago

The rate constant for this second‑order reaction is

o-na [289]

Answer:

daddadaddddddddddddddddddddddddddddda

Explanation:

dddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddd

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3 years ago

Which net ionic equation best represents the reaction that occurs when an aqueous solution of barium chloride is mixed with an a

vovangra [49]

Answer : The net ionic equation will be:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The balanced molecular equation will be,

$BaCl_2(aq)+Li_2SO_4(aq)\rightarrow 2LiCl(aq)+BaSO_4(s)$

The complete ionic equation in separated aqueous solution will be,

$Ba^{2+}(aq)+2Cl^-(aq)+2Li^+(aq)+SO_4^{2-}(aq)\rightarrow 2Li^+(aq)+2Cl^-(aq)+BaSO_4(s)$

In this equation the species, $Li^+\text{ and }Cl^{-}$ are the spectator ions.

By removing the spectator ions , we get the net ionic equation.

The net ionic equation will be:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

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3 years ago

The elements found in Group 2A, from top to bottom, include beryllium, magnesium, calcium, strontium, and barium. Which of these

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Answer is beryllium.

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3 years ago

The half life of an isotope X is 3 days how long will it take for 1/2 of a 10 G sample to decay

lorasvet [3.4K]

Answer: It will take 3 days for half of a 10 g sample to decay.

Explanation:

Half-life of sample of an isotope X = 3 days

$\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{3 days}=0.231 day^{-1}$