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Which formula equation represents the burning of sulfur to produce sulfur dioxide?

Leni [432]

Answer:

S(s) + O2(g) --> SO2(g)

Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

Explanation:

The reaction is given as;

Sulfur + oxygen --> Sulphur dioxide

Sulphur = S

Oxygen = O2

Sulfur dioxide = SO2

So we have;

S(s) + O2(g) --> SO2(g)

The crrect option is option A. Upper S (s) plus upper O subscript 2 (g) right arrow with delta above upper S upper O subscript 2 (g).

5 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Express the rate of reaction in terms of the change in concentration of each of the reactants and p

Tanzania [10]

Answer : The [H] is increasing at the rate of 0.36 mol/L.s

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.

The given rate of reaction is,

$2D(g)+3E(g)+F(g)\rightarrow 2G(g)+H(g)$

The expression for rate of reaction :

$\text{Rate of disappearance of }D=-\frac{1}{2}\frac{d[D]}{dt}$

$\text{Rate of disappearance of }E=-\frac{1}{3}\frac{d[E]}{dt}$

$\text{Rate of disappearance of }F=-\frac{d[F]}{dt}$

$\text{Rate of formation of }G=+\frac{1}{2}\frac{d[G]}{dt}$

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

$\text{Rate of reaction}=-\frac{1}{2}\frac{d[D]}{dt}=-\frac{1}{3}\frac{d[E]}{dt}=-\frac{d[F]}{dt}=+\frac{1}{2}\frac{d[G]}{dt}=+\frac{d[H]}{dt}$

Given:

$-\frac{d[D]}{dt}=0.18mol/L.s$

As,

$-\frac{1}{2}\frac{d[D]}{dt}=+\frac{d[H]}{dt}=0.18mol/L.s$

and,

$+\frac{d[H]}{dt}=2\times 0.18mol/L.s$

$+\frac{d[H]}{dt}=0.36mol/L.s$

Thus, the [H] is increasing at the rate of 0.36 mol/L.s

5 0

3 years ago

Hi, can someone help me with chemistry

Dominik [7]

Idk look it up on another website

6 0

3 years ago

For the reaction shown, calculate how many grams of each product form when the following amounts of reactant completely react to

stira [4]

This may help you
<span>You need to use some stoichiometry here. The only way to do that is if you're working in moles. Since you're given grams of Al, you can convert that moles by dividing by the molar mass. Then from looking at the coefficients in your equation, you can see that for however many moles of Al react, the same numbers of moles of Fe will be produced, but only half as many moles of Al2O3 will be produced. To go back to grams, multiply the moles of each product that you get by their molar masses!</span>

7 0

3 years ago

1) If the pressure on 2.6 liters of a gas at 860 Torr is increased to 1000 Torr,

LenKa [72]

2.24 liters is the volume of the gas if pressure is increased to 1000 Torr.

Explanation:

Data given:

Initial volume of the gas V1 = 2.6 liters

Initial pressure of the gas P1 = 860 Torr 1.13 atm

final pressure on the gas P2 = 1000 Torr 1.315 atm

final volume of the gas after pressure change V2 =?

From the data given above, the law used is :

Boyles Law equation:

P1V1 = P2V2

V2 = P1V1/P2

= 1.13 X 2.6/ 1.31

= 2.24 Liters

If the pressure is increased to 1000 Torr or 1.315 atm the volume changes to 2.24 liters. Initially the volume was 2.6 litres and the pressure was 860 torr.

6 0

3 years ago