All categories

3 years ago

How many grams of FeCl 3 are in 250. mL of a 0.100 M solution?

Chemistry

1 answer:

Naddika [18.5K]3 years ago

5 0

Answer:

The correct answer is option B

Explanation:

$$Molarity=\frac{Weight}{Molecular \,weight} \frac{1000}{V(in \, ml)}$

Given values,

Molarity of $FeCl_3=0.100M$

Volume of solution, $V=250ml$

Molecular weight of $FeCl_3=162.2$

Substituting this values in Molarity formula, we get

$0.1=\frac{weight}{162.2} \times\frac{1000}{250} $\\$\Rightarrow 16.22=weight\times4$\\$\Rightarrow weight=\frac{16.22}{4} $\\$\therefore weight=4.06g$

You might be interested in

There are two open cans of soda on the table. One can was just taken from the refrigerator and the other was taken from the cupb

nika2105 [10]

I think the correct answer from the choices listed above is option C. The can from the cupboard will lose carbon dioxide more quickly because it is warmer and gases are less soluble in warmer temperatures. Solubility of gases is a strong function of temperature and as well as pressure.

5 0

3 years ago

Read 2 more answers

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

Check out this app! It's millions of students helping each other get through their schoolwork. https://brainly.app.link/qpzV02Ma

marishachu [46]

Answer:

why did you post a link to brainly when we are on brainly already??????

Explanation:

8 0

2 years ago

Viết các đồng phân cấu tạo mạch hở của C4H6O2 cùng nhóm chức axit

Aleksandr-060686 [28]

Answer:

+ axit

CH2=CH-CH2-COOH,

CH3-CH=CH-COOH (tính cả đồng phân hình học)

CH2=C(CH3)-COOH.

+ este

HCOOCH=CH-CH3 (tính cả đồng phân hình học)

HCOO-CH2-CH=CH2,

HCOOC(CH3)=CH2.

CH3COOCH=CH2

CH2=CH-COOCH3

8 0

3 years ago

Calculate the volumeof 1.0M NaOH necessary to completely neutralize 100mL of 0.50M of phosphoric acid (H3PO4)?

blondinia [14]

Answer:

50 mL

Explanation:

In case of titration , the following formula is used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.50M

V₁ = 100 mL

M₂ = 1.0M

V₂ = ?

Using the above formula , the volume of base , can be calculated as ,

M₁V₁ = M₂V₂

substituting the respective values ,

0.50M * 100 mL = 1.0M * V₂

V₂ = 50 mL

3 0

3 years ago