What two kinds of energy are associated with flames

Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2

lapo4ka [179]

Answer: Concentration of $CO$ = 0.328 M

Concentration of $Cl_2$ = 0.328 M

Concentration of $COCl_2$ = 0.532 M

Explanation:

Moles of $CO$ and $Cl_2$ = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of $CO$ and $Cl_2$ = $\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M$

The given balanced equilibrium reaction is,

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

Initial conc. 0.860M 0.860M 0

At eqm. conc. (0.860-x) M (0.860-x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[COCl_2]}{[CO][Cl_2]}$

Now put all the given values in this expression, we get :

$4.95=\frac{x}{(0.860-x)^2}$

By solving the term 'x', we get :

x = 0.532 M

Thus, the concentrations of $CO,Cl_2\text{ and }COCl_2$ at equilibrium are :

Concentration of $CO$ = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of $Cl_2$ = (0.860-x) M = (0.860-0.532) M = 0.328 M

Concentration of $COCl_2$ = x M = 0.532 M

3 0

3 years ago

If hydrofluoric acid is a stronger acid than acetic acid, which statement is most likely true?

Alchen [17]

Answer: The statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.

Explanation:

A strong acid upon dissociation gives a weak conjugate base. This can also be said as stronger is the acid, weaker will be its conjugate base or vice-versa.

Hydrofluoric acid is a strong base as it dissociates completely when dissolved in water.

For example, $HF \rightleftharpoons H^{+} + F^{-}$

The conjugate base is $F^{-}$ which is a weak base.

Acetic acid is a weak acid as it dissociates partially when dissolved in water. So, the conjugate base of acetic acid is a strong base.

$CH_{3}COOH \rightarrow CH_{3}COO^{-} + H^{+}$

Thus, we can conclude that the statement conjugate base of hydrofluoric acid is weaker than that of acetic acid is most likely true.

4 0

3 years ago

Atoms join together to form compounds bt gaining,loosing or sharing

user100 [1]

Answer:

the answer is C) sharing

Explanation:

positive ions & negative ions form when atom s lose or gain electrons. Covalent bonds form when atoms share electrons.Metallic bonds form by the attraction of metal ions and the electrons around them. Covalent compounds form when atoms of elements share electrons.

4 0

3 years ago

Read 2 more answers

The balanced redox reactions for the sequential reduction of vanadium are given below.

Minchanka [31]

Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

$2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)$ $2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)$

$2V^{3+} (aq)+ Zn (s)\rightarrow 2V^{2+}(aq)+Zn^{2+}(aq)$

On adding above equations, we get the following combined equation.

$2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)$

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

$12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}$

From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.

Let us use this as a conversion factor to find the moles of Zn.

$0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn$

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

$0.00059895mol Zn\times \frac{65.38gZn}{1molZn}=0.0392gZn$

We need 0.0392 grams of Zn metal to completely reduce vanadium.

6 0

3 years ago

N2H4 + N2O4 --> N2 + H2O

ahrayia [7]

Answer:

The limiting reagent is N2O4
14,09g

Explanation:

First, we adjust the reaction.

$2N_{2} H_{4}$ + $N_{2} O_{4}$ ⇄ $6N_{2} + 4H_{2}O$

Second, we assume that the participating moles are equal to the stoichiometric ratios because we do not know the amounts of the reagents.

We can determinate what is the limiting reagent comparing of product amounts which can be formed from each reactant.

Using $N_{2} H_{4}$ to form $H_{2}O$

$molH_{2} O = 1mol N_{2} H_{4} } . \frac{4 mol H_{2} O}{2mol N_{2} H_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}H_{4} }{32,04\frac{g}{mol} N_{2} H_{4} }$

$molH_{2} O = 1, 125 mol$

Using $N_{2} O_{4}$ to form $H_{2} O$

$molH_{2} O = 1mol N_{2} O_{4} } . \frac{4 mol H_{2} O}{1mol N_{2} O_{4} }. \frac{18\frac{g}{mol} H_{2} O}{1mol H_{2} O_} } . \frac{ 1 mol N_{2}O_{4} }{92\frac{g}{mol} N_{2} O_{4} }$

$molH_{2} O = 0,783 mol$

The limiting reagent is N2O4, because can produce only 0, 783 mol of H2O.

This is the minimum measure can be formed of each product.

∴ $MassOfH_{2}O = 0,783mol . 18\frac{g}{mol}$

$MassOfH_{2}O = 14,09g$

5 0

3 years ago