What color offspring did you get when you crossed a brown "BB" rabbit to a white "bb" rabbit?

Consider the elements in the periodic table. The stair-step line between the pink squares and the yellow squares separates the _

photoshop1234 [79]

The choices for this are as follows:

A) gases; solids
B) metals; nonmetals
C) nonmetals; metals
<span>D) reactive; nonreactive
</span>
I think the correct answer is option B. The stair-step line between the pink squares and the yellow squares separates the metals from the nonmetals. Hope this helps.

6 0

2 years ago

Compound X has a molar mass of 266.64 g/mol and the following composition: aluminum 20.24% chlorine 79.76% Write the molecular f

N76 [4]

Answer:

Explanation:

Assume we have 100g of this substance. That means we would have 20.24g of Cl and 79.76g of Al. Now we can find how many moles of each we have:

$\frac{79.76 \:g}{35.45 \: g/mol}$ = 2.25 mol of chlorine

$\frac{20.24 \: g}{26.98 \: g/mol}$ = 0.750 mol of Al.

To form a integer ratio, do 2.25/0.75 = 2.99999 ~= 3.

So the ratio is essentially Al : Cl => 1 : 3. To the compound is possibly $AlCl_3$ .

However, it says it has a molar mass of 266.64 g/mol, and since AlCl3 has a molar mass of 133.32, it must be $Al_2Cl_6$ .

Actually this molecule isn't exactly AlCl3 (which is ionic). Al2Cl6 forms a banana bond where Cl acts as a hapto-2 ligand. But that's a bit advanced. All you need to know is X = Al2Cl6

5 0

2 years ago

Helpppppppppppppppppppppppppppp

worty [1.4K]

Answer:

evaporation or water runoff that's my tips

rain or thunder storm

6 0

2 years ago

Read 2 more answers

One mole of titanium contains how many atoms

vitfil [10]

6.022 x 10 23 titanium atoms

In 47.88 grams of titanium, there is one mole, or 6.022 x 1023 titanium atoms.

Hope this helps :)

8 0

3 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

B)

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

3 0

3 years ago