All categories

3 years ago

Sulfur reacts with oxygen to produce sulfur trioxide gas. Balance the equation. S+(sub2) -----> __SO(sub3)

1 answer:

6 0

2S + 3O2 ==> 2SO3
Let's start off with the unbalanced equation and count atoms of each element on both sides of the arrow.
S + O2 ==> SO3
There's 1 sulfur atom on both sides which is good. However, there's 2 oxygen atoms on the left and 3 on the right. So calculate the Least common multiple (LCM) of 2 and 3 which is 6. And adjust the coefficients on both sides to have 6 oxygen atoms on each side, giving:
S + 3O2 ==> 2SO3
Now we have 6 oxygen atoms on each side which is good, but there's only 1 sulfur atom on the left and 2 on the right. Since the LCM of 1 and 2 is 2, all we need to do is adjust the coefficient on the left to get 2 sulfur atoms there. So:
2S + 3O2 ==> 2SO3
We now have 2 sulfur atoms on each side of the equation and 6 oxygen atoms on each side, so the resulting equation is balanced.

You might be interested in

What is the first step in creating a unit conversion factor?

NNADVOKAT [17]

Determining the units.

5 0

3 years ago

Show your calculation by uploading a picture. Calculate the molar mass of ammonia, NH3

Cloud [144]

Answer:

17.04 g/mol

Explanation:

Molar Mass of NH₃

we know that

Nitrogen has 14.01 gram/mol

And Hydrogen has 1.01 gram/mol

but we have 3 Hydrogens So we multiply

1.01 by 3 i.e., 3.03

Now, add

14.01

+ 3.03

17.04

So, The molar mass of ammonia, NH₃ is

17.04 g/mol

-TheUnknownScientist

5 0

3 years ago

Help plssss The question is in the picture above

Paul [167]

Answer:

The answer is : A

8 0

3 years ago

Read 2 more answers

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

Answer: The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago

There are some exceptions to the trends of first and successive ionization energies. For each of the following pairs, explain wh

NNADVOKAT [17]

Answer:

(b) IE₂ of Ga > IE₂ of Ge

Explanation:

Electronic configuration of Ga is [Ar] 3d¹⁰4s²4p¹

Electronic configuration of Ge is [Ar] 3d¹⁰4s²4p²

After 1st ionisation , Ga becomes [Ar] 3d¹⁰4s² and becomes stable . Its

2 nd ionisation requires higher amount of ionisation energy. In case of Ge , there are 2 electrons in outermost orbital so it becomes stable after ionisation of 2 electrons.

4 0

3 years ago

Read 2 more answers

Sulfur reacts with oxygen to produce sulfur trioxide gas. Balance the equation. __S+__(sub2) -----> __SO(sub3)

Sulfur reacts with oxygen to produce sulfur trioxide gas. Balance the equation. S+(sub2) -----> __SO(sub3)