Number of different bond angles for pbr2f3

GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

$2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}$

$\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}$

$\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}$

$\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}$

$\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}$

Based on no. of iron reacted,

$\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)$

n = m/M

$\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}$

% composition of $FeCl_3$

= $(9.75 / 10.39)^{*} 100$

= 94%

6 0

3 years ago

How many grams of a 23.4% by mass naf solution is needed if you want to have 1.33 moles of naf?

jenyasd209 [6]

Percentage by mass is the mass of NaF present in 100 g of the solution.
the percentage by mass of NaF is 23.4 %
this means that in 100 g of solution, mass of NaF present is 23.4 g
the number of moles of NaF present - 1.33 mol
mass of NaF - 1.33 mol x 42 g/mol = 55.9 g
when there's 23.4 g of NaF - mass of solution is 100 g
therefore when there's 55.9 g of NaF - mass of solution is 100 / 23.4 x 55.9
= 239 g
mass of solution required is 239 g

6 0

4 years ago

________ properties can be observed only when the substance in a sample of matter are changed into different substances

RUDIKE [14]

I believe it's Chemical properties

5 0

3 years ago

What are the concentrations of Cu2+, NH3, and Cu(NH3)42+ at equilibrium when 18.8 g of Cu(NO3)2 is added to 1.0 L of a 0.800 M s

Svetllana [295]

Answer:

Explanation:

Cu(NO₃)₂ + 4NH₃ = Cu(NH₃)₄²⁺ + 2 NO₃⁻

187.5 gm 4M 1 M

187.5 gm reacts with 4 M ammonia

18.8 g reacts with .4 M ammonia

ammonia remaining left after reaction

= .8 M - .4 M = .4 M .

187.5 gm reacts with 4 M ammonia to form 1 M Cu(NH₃)₄²⁺

18.8 g reacts with .4 M ammonia to form 0.1 M Cu(NH₃)₄²⁺

At equilibrium , the concentration of Cu²⁺ will be zero .

concentration of ammonia will be .4 M

concentration of Cu(NH₃)₄²⁺ formed will be 0.1 M

3 0

3 years ago

An erlenmeyer flask contains 15.00mL of 0.030 M HCI before titration. 5.00 mL of 0.050 of M NaOH is added to the HCI in the flas

e-lub [12.9K]

Answer:

1.1

0.050*0.005

0.00025

Explanation:

8 0

3 years ago