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Flauer [41]
3 years ago
9

Oxidation number of au in kaucl4

Chemistry
2 answers:
scZoUnD [109]3 years ago
8 0

+7 is the correct answer for this.

DanielleElmas [232]3 years ago
3 0
KauCl4 :

K = + 1 

au = + 7

Cl = - 2

hope this helps!

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Phosphate is derived from the titration of phosphoric acid, H3PO4. The three pKa’s of H3PO4 are 2.15, 7.20, and 12.35 respective
aleksandr82 [10.1K]

Answer:

Answers are in the explanation

Explanation:

The equilibriums corresponding for each ionization are:

H₃PO₄ ⇄ H₂PO₄⁻ + H⁺ pka: 2,15

H₂PO₄⁻ ⇄ HPO₄²⁻ + H⁺ pka: 7,20

HPO₄²⁻ ⇄ PO₄³⁻ + H⁺ pka: 12,35

At pH 5,7 all H₃PO₄ reacts to produce H₂PO₄⁻ and the consumed H₂PO₄⁻ to form HPO₄²⁻ is not relevant yet. Thus, dominant for is <em>H₂PO₄⁻</em>

I hope it helps!

5 0
4 years ago
The equation with the concentrations of the products of a reaction raised to their stoichiometric coefficients divided by the co
liberstina [14]

Explanation:

This is the description of the equilibrium constant equation. For example:

Given a generic reaction: aA + bB \longleftrightarrow cC+ dD

Where A.B,C and D are the reactants and products and a,b,c and d are their stoichiometric coefficients. The equilibrium constant of this reaction will be:

K=\frac{[C]^c*[D]^d}{[A]^a*[B]^b}

<em>Note: is important to have in mind that the substances in solid or liquid (pure) state should not be included in the K equation.</em>

6 0
4 years ago
A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista
baherus [9]

Explanation:

Formula for work done is as follows.

           W = -k \frac{q_{1}q_{2}}{d}    

where,  k = proportionality constant = 8.99 \times 10^{9} Jm/C^{2}

            q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C

            q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C

            d = separation distance = 0.45 nm = 0.45 \times 10^{-9} m

Now, we will put the given values into the above formula and calculate work done as follows.

         W = -k \frac{q_{1}q_{2}}{d}    

           = \frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}  

           = 3.68 \times 10^{-18} J

Thus, we can conclude that work required to increase the separation of the two ions to an infinite distance is 3.68 \times 10^{-18} J.

7 0
3 years ago
Please help me with this​
elena-s [515]

Answer:

5

Explanation:

5 0
3 years ago
1500 millimeters to km
gizmo_the_mogwai [7]
0.0015 kilometers is for sure the answer!
4 0
3 years ago
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