Name this compoundNa[Au(CN)4]

Chemistry

1 answer:

trapecia [35]3 years ago

4 0

Answer:

the only remote information i could find about this element was sodium

dicyanidoaurate(I)

sodium dicyanoaurate(1-)

Na[Au(CN)2]

sodium dicyanidoaurate(1-)

but as you see these go towards Na[Au(CN)2] not Na[Au(CN)4], so I will leave the rest up to you.

Have a good day.

Explanation:

https://pubchem.ncbi.nlm.nih.gov/compound/5460536

that is the website i found that off of

https://www.webqc.org/molecular-weight-of-Na%5BAu%28CN%294%5D.html

this one tells you the element names.

You might be interested in

Given the reaction below:

lidiya [134]

The given equation is is

$2Cu+8HNO_{3}-->3Cu(NO_{3})_{2}+2NO+4H_{2}O$

this is not balanced

The balanced equation will be:

$3Cu+8HNO_{3}-->3Cu(NO_{3})_{2}+2NO+4H_{2}O$

Now as per it if we are using eight moles of HNO3 we are getting two moles of NO

so for each mole of NO we have to take four moles of HNO3

Hence the mole ratio of NO produced to HNO3 reacted will be = 1:4

4 0

3 years ago

Using your value of Ksp, and starting with an equilibrium system consisting of a saturated solution of calcium hydroxide, predic

Ugo [173]

Answer:

(A) $HNO_{3}$ Solubility will increase.

(B) $NaOH$ Solubility will decrease.

(C) $Ca(NO_{3} )_{2}$ Solubility will decrease.

Explanation:

a) According to Le Chatelier's Principle when a change is introduced in a reaction system at equilibrium, the system responds by the reaction shifting in the direction to counter the change introduced.

When the change introduced is addition of acid, some $OH^{-}$ are consumed due to it and the system would respond by reaction shifting towards creation of more $OH^{-}$ in the system by increased dissolution. Thus solubility increase.

(b) When $OH^{-}$ are added, the reaction shifts to counter the change introduced of increased $OH^{-}$ by shifting to side that decreases the increased $OH^{-}$ . That happens by reaction shifting towards some $Ca (OH)_{2}$ precipitating back. Hence solubility decreases.

(c) The change introduced is addition of $Ca^{2+}$ which will be countered by reaction shifting to side that decreases increased $Ca ^{2+}$ , which happens by some $Ca(OH)_{2}$ precipitating back. Hence solubility decreases.