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3 years ago

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Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of t

he following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 4NO2 + 6H2O 16 Δ[H2O]Δt 14 Δ[NO2]Δt - 14 Δ[NH3]Δt - 17 Δ[O2]Δt All of the above are valid expressions of the reaction rate.

1 answer:

Veseljchak [2.6K]3 years ago

3 0

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

$4NH_3+7O_2\rightarrow 4NO_2+6H_2O$

The expression for rate of reaction for the reactant :

$\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}$

$\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}$

The expression for rate of reaction for the product :

$\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}$

From this we conclude that, all the options are correct.

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OverLord2011 [107]

Because the physical appearance change hope this helps
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5 0

3 years ago

Read 2 more answers

In the laboratory, a student dilutes 18. 9 ml of a 10. 0 m perchloric acid solution to a total volume of 250. 0 ml. what is the

Sever21 [200]

The concentration of diluted solution is 0.756M.

From the question given above, the following data were obtained:

Volume of stock solution (V1) = 18.9 mL

Molarity of stock solution (M1) = 10 M

Volume of diluted solution (V2) = 250 mL

Molarity of diluted solution (M2) =?

We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:

M1V1 = M2V2

10 × 18.9 = M2 ×250

189 = M2 × 250

Divide both side by 100

M2 = 189 / 250

M2 = 0.756 M

Therefore, the molarity of the diluted solution is 0.756 M.

Thus the concluded that concentration of the dilute acid is 0.756 M.

Learn more about concentration of diluted solution: brainly.com/question/10725862

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7 0

1 year ago

The analysis of a hydrocarbon revealed that it was 85.7% C and 14.3% H by mass. When 1.77 g of the gas was stored in a 1.500-L f

gtnhenbr [62]

Answer:

The formula of hydrocarbon = $C_3H_6$

Explanation:

$Moles =\frac {Given\ mass}{Molar\ mass}$

% of C = 85.7

Molar mass of C = 12.0107 g/mol

% moles of C = 85.7 / 12.0107 = 7.14

% of H = 14.3

Molar mass of H = 1.00784 g/mol

% moles of H = 14.3 / 1.00784 = 14.19

Taking the simplest ratio for C and H as:

7.14 : 14.19 = 1 : 2

The empirical formula is = $CH_2$

Also, Given that:

Pressure = 508 Torr

Temperature = 17 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (17 + 273.15) K = 290.15 K

Volume = 1.500 L

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 62.3637 L.torr/K.mol

Applying the equation as:

508 Torr × 1.500 L = n × 62.3637 L.torr/K.mol × 290.15 K

⇒n = 0.0421 moles

Given that :

Amount = 1.77 g

Molar mass = ?

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.0421\ moles= \frac{1.77\ g}{Molar\ mass}$

Molar mass of the hydrocarbon = 42.04 g/mol

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12 + 2×1= 14 g/mol

Molar mass = 42.04 g/mol

So,

Molecular mass = n × Empirical mass

42.04 = n × 14

⇒ n = 3

<u>The formula of hydrocarbon = $C_3H_6$ </u>

5 0

3 years ago

Malonic acid, h2c3h2o4, is a diprotic acid with ka1= 1.5 x 10-3and ka2= 2.0 x 10-6. what is the concentration of the malonate an

Bess [88]

In a 0.15 m solution of malonic acid, the amount of the malonate anion, c3h2o42-, is $1.68*10^{-4} M$ .

Explanation:

For the first equlibria;

initial concentration of $H_{2}C_{3}H_{2}OH=0.15$

initial concentration of $HC_{3}H_{2}O_{4}^{-} =0$

initial concentration of $H^{+} =0$

Let concentration change of $H_{2}C_{3}H_{2}OH=-x$

Let concentration change of $HC_{3}H_{2}O_{4}^{-} =+x$

Let concentration change of $H^{+} =+x$

Let concentration in equilibrium state of $H_{2}C_{3}H_{2}OH=0.15-x$

Let concentration in equilibrium state of $HC_{3}H_{2}O_{4}^{-} =x$

Let concentration in equilibrium state of $H^{+} =x$

Therefore, $Ka_{1}=\frac{x*x}{0.15-x}=\frac{1.5*10^{-3}}{1} =\frac{x^{2}}{0.15-x}$

$x^{2}+0.0015x-0.000225=0$

Which is a quadratic equation in x, solving it gives;

$x=0.01427$

Therefore, $HC_{3}H_{2}O_{4}^{-} =x= 0.01427M$

Now for the second equlibria;

initial concentration of $HC_{3}H_{2}O_{4}^{-}=0.01427$

initial concentration of $C_{3}H_{2}O_{4}^{2-} =0$

initial concentration of $H^{+} =0$

Let concentration change of $HC_{3}H_{2}O_{4}^{-}=-x$

Let concentration change of $C_{3}H_{2}O_{4}^{2-} =+x$

Let concentration change of $H^{+} =+x$

Let concentration in equilibrium state of $HC_{3}H_{2}O_{4}^{-}=0.01427-x$

Let concentration in equilibrium state of $C_{3}H_{2}O_{4}^{2-} =x$

Let concentration in equilibrium state of $H^{+} =x$

Therefore, $Ka_{2}=\frac{x*x}{0.01427-x}=\frac{2.0*10^{-6}}{1} =\frac{x^{2}}{0.01427-x}$

$2.854*10^{-8}-(2.0*10^{-6})x=x^{2}$

Therefore, $C_{3}H_{2}O_{4}^{2-} =x= 1.68*10^{-4}M$

<h3>What is a malonic acid?</h3>

The chemical formula of malonic acid is CH2(COOH)2. Malonates include the ionized form of malonic acid as well as its esters and salts. For instance, the diethyl ester of malonic acid is called diethyl malonate.

Polyester and polymers both employ malonic acid as a precursor. Both the scent business and the flavoring sector utilize it. It is employed to regulate acidity. Malonic acid contributes to the production of fatty acids in people. Malonic acid has been found in a number of foods outside of the human body, including red beets, corn, scarlet beans, common beets, and cow milks, although its presence has not been defined. The group of organic substances known as dicarboxylic acids and derivatives includes malonic acid, also known as malonate or H2MALO.

To know more about malonic acid, visit:

brainly.com/question/13943912?referrer=searchResults

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4 0

1 year ago

A sample of 9.00 grams of aluminum metal is added to an excess of hydrochloric acid. The volume of hydrogen gas produced at stan

marysya [2.9K]

Answer:

The volumen of hydrogen (H) is 11.22 L

Explanation:

The balanced reaction between the aluminum (Al) and the hydrochloric acid (HCl) is:

2Al(s) + 6HCl(l) --> 2AlCl3(ac) + 3H2(g)

It is a reduction-oxidation reaction.

At the beginning we have 9.00 grams of Al what represents an certain amount of moles. Then, we know the molar mass of Al is 26.9815 g/mol, so the moles content in 9.00 g of Al are :

9.00 g Al * (1 mol Al / 26.9815 g Al)= 0.33356 mol Al

Now, we have to conisdered the molar relation between the Al and H2, according to the balanced reaction performed above. That is 2:3 (Al:H2)

Then,

0.33356 mol Al * (3 mol H2/ 2mol Al) = 0.50034 mol H2

In this point, we can considered the H2 like a noble gas, because the question ask for standard temperature and pressure (0° C= 273.15K and 1 atm).

Let us remember the noble gases equation

PV=nRT

P: pressure

V: Volumen

n: number of moles

R: gases constant

T: temperature

For the volumen, the equation is:

V= (nRT/P)

Now: V= (0.50034 mol * 0.0821 (L*atm/K*mol) * 273.15 K)/ (1 atm)

V= 11.22 L

3 0

3 years ago