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3 years ago

9

Sterling silver is a mixture of the metals silver and copper.It is used to make jewellery.sometimes sterling silver causes green

staining on skin.Which metal do you think is responsible for this?Explain why.

1 answer:

slavikrds [6]3 years ago

8 0

The composition of 925 Sterling Silver lends itself to the occasional green discoloration because of the presence of copper

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1) How many moles are in 4.0x10^24 atoms?

-Dominant- [34]

Answer:

<h2>6.64 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{4 \times {10}^{24} }{6.02 \times {10}^{23} } \\ = 6.644518...$

We have the final answer as

<h3>6.64 moles</h3>

Hope this helps you

3 0

3 years ago

HELP ASAP ILL GIVE BRAINIEST!!!

UNO [17]

Answer:

1. the group number of sodium is 1 and it is a metal

2. the group number of helium is 18 and it is a nonmetal

3. the group number of iodine is 17 and it is a nonmetal

4.the group number of calcium is 2 and it is a metal

5. lithium has similar properties to potassium

6. calcium has similar properties to magnesium

7. neon has similar properties to xenon

8. Iodine has similar properties to chlorine

5 0

3 years ago

A young girl has blue eyes.

Leviafan [203]

Answer:

D

Explanation:

( I hope that this helps )

4 0

3 years ago

Read 2 more answers

SeF6 1. Lewis Structure 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs

Anarel [89]

The answer is- $SF_{6}$ is octahedral in electronic and molecular geometry with 6 Fluorine atoms bonded to central atom S.

Lewis structures are the diagrams in which the valence electrons of the atoms of a compound are arranged around the atoms showing the bonding between the atom and the lone pair of electrons existing in the molecule.

Determine the molecular geometry of $SF_{6}$ .

Valence Shell Electron Pair Repulsion theory is commonly known as VSEPR theory and it helps to predict the geometry of molecules.
According to this theory, electrons are arranged around the central atom of the molecule in such a way that there is minimum electrostatic repulsion between these electrons.
Now, calculate the total number of valence electrons in $SF_{6}$ .

$Valence\ electrons\ in\ SF_{6}= Valence\ electrons\ in\ S +\ 6(Valence\ electrons\ in\ F)$

Valence electrons of S = 6

Valence electrons of F = 7

Thus, the valence electrons in $SF_{6}$ are-

$Valence\ number\ of\ electrons\ in\ SF_{6} = (6) + 6(7) = 48\ electrons.$

The Lewis structure of $SF_{6}$ is - (Image attached).
In the structure, the number of atoms bonded to central atom (S) = 6.
Number of non-bonding electron pairs on the central atom = 0 (as all the valence electrons are bonded to F).
Electronic geometry in case of 6 bond pairs is octahedral.
Molecular geometry us also octahedral with bond angles 90°.
Central atom is sp3d2 hybridised.
$SF_{6}$ is a non-polar molecule.

To learn more about Lewis structures visit:

brainly.com/question/12307841?referrer=searchResults

#SPJ4

7 0

1 year ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago