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3 years ago

Liters of a 2.40 M

Chemistry

1 answer:

kotykmax [81]3 years ago

6 0

Answer:

1.42 L

Explanation:

Step 1:

The following data were obtained from the question :

Molarity of KBr = 2.40 M

Mole of KBr = 3.40 moles

Volume of solution =?

Step 2:

Determination of the volume of the solution.

Molarity of solution is simply the mole of the solute per unit volume the of solution. It is given as :

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 3.4/2.4

Volume = 1.42 L

Therefore, the volume of the solution is 1.42 L

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If my aunt means degree Celsius or degrees Fahrenheit

JulijaS [17]

If it is 60 Celsius that would conver to fare height by means of this equation; (1.8*60)+32°F
Which would come out to.... 140° Fahrenheit... Hardly seems like chilly conditions.

7 0

3 years ago

Be sure to answer all parts. Write the balanced equations corresponding to the following rate expressions: a) rate = − 1 3 Δ[CH4

Alinara [238K]

Answer : The balanced equations will be:

(a) $3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $2N_2O_5\rightarrow 2N_2+5O_2$

(c) $2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

(a) $Rate=-\frac{1}{3}\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[H_2O]}{dt}=-\frac{d[CO_2]}{dt}=+\frac{1}{4}\frac{d[CH_3OH]}{dt}$

The balanced equations will be:

$3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH$

(b) $Rate=-\frac{1}{2}\frac{d[N_2O_5]}{dt}=+\frac{1}{2}\frac{d[N_2]}{dt}=+\frac{1}{5}\frac{d[O_2]}{dt}$

The balanced equations will be:

$2N_2O_5\rightarrow 2N_2+5O_2$

(c) $Rate=-\frac{1}{2}\frac{d[H_2]}{dt}=-\frac{1}{2}\frac{d[CO_2]}{dt}=-\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2CO_3]}{dt}$

The balanced equations will be:

$2H_2+2CO_2+O_2\rightarrow 2H_2CO_3$

4 0

3 years ago

Molten zinc chloride can be electrolysed.

Crank

Answer: The products of electrolysis are Zn(l) and $Cl_2$ gas

Explanation:

Electrolysis of a subastance is breaking it into its constituents by the action of electrical current.

In the electrolysis of molten zinc chloride, zinc metal is produced at the cathode which is a negative electrode and chlorine gas produces as the anode which is a positive electrode.

$ZnCl_2(l)\rightarrow Zn^{2+}+2Cl^-$