25 grams of HF is how many moles

What do cubic measurements, live cm3, mesure?"<br> O length<br> volume

anygoal [31]

The answer is volume.

6 0

3 years ago

Analyze your results.

raketka [301]

#b

According to Le C ha.te llors principle of we increase concentration of reactants or products equilibrium shifts.

Yes increases

#c

Rate of reaction also increases

#e

Stated in b

3 0

2 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

The element in Group 14 Period 4 has how many neutrons and is it a metal, nonmetal, metalloid or noble gas.

monitta

Number of neutrons=41, and it is metalloid

<h3>Further explanation</h3>

Group 14⇒valence electron = 4(ns²np²)

Period 4⇒n=4

So electron configuration of the element :

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p² = 32 electron=atomic number

The element with atomic number 32, which is in period 4 and group 14 is Ge-Germanium

There are seven elements that can be classified as metalloids, namely boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po).

While the mass number: 73

So the number of neutrons = mass number-atomic number

$\tt 73-32=41$

8 0

3 years ago

A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 234 picometers and its

AleksAgata [21]

Answer:

$\delta=101.13 g/cm^3$

Explanation:

As can be seen in the Figure in a face-centered cubic unit cell you have:

Six halves of atoms
Eight 1/8 of atom (1 in each corner)

In total:

$n_{atom}=6*0.5+8*\frac{1}{8}=4 atoms$

Now, each side of the cell is 234 picometers (2.34e-8 cm) long

$V_{cell}=L^3=(2.34e^{-8} cm)^3$

$V_{cell}=1.28*10^{-23}cm^3$

Atoms per cm3:

$n=\frac{4 atoms}{1.28*10^{-23}cm^3}$

$n=3.12*10^{23} atoms/cm^3$

Expressing in mass:

$\delta=3.12*10^{23} atoms/cm^3* \frac{1 mol}{6.022*10^{23}}*195.08 g/mol$

$\delta=101.13 g/cm^3$

6 0

4 years ago