Can you please help me

Two radius of an atom is equal to the diameter. Adding up all the diameter of the atoms, it should be equal to 9.5 mm. Therefore, we simply convert the units to the same units then divide 1.35 A to 9.5 mm. We calculate as follows:

no. of atoms = 0.0095 m / 1.35x10^-9 m = 7037037 atoms

Hope this answers the question. Have a nice day.

6 0

3 years ago

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A 1.0-L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys the buffer? (a) adding 0.050 mol of HCl (b) ad

Volgvan

Answer:

(a) adding 0.050 mol of HCl

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.

In the buffer:

1.0L × (0.10 mol / L) = 0.10 moles of HF -Weak acid-

1.0L × (0.050 mol / L) = 0.050 moles of NaF -Conjugate base-

-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-

Thus:

(a) adding 0.050 mol of HCl: The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid destroying the buffer.

(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer wouldn't be destroyed.

(c) adding 0.050 mol of NaF: The addition of conjugate base doesn't destroy the buffer

3 0

3 years ago

A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but additio

zhannawk [14.2K]

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate $SO_4~^-^2$ and the chloride $Cl^-$ :

Sulfate

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.( $Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2$ ), which are NOT soluble.

Chloride

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ( $Pb^+^2~,Ag^+~,Hg_2~^+^2$ ), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I), lead" and our possibilities are:

"Calcium, strontium, or barium".

I hope it helps!

7 0

3 years ago