someone sprays perfume in another room. you smell it a few minutes later. what process is this an example of

How many moles of O2 are needed to burn 2.56 moles of CH3OH?

lukranit [14]

Answer:

$n_{O_2}=3.84molO_2$

Explanation:

Hello!

In this case, since the combustion reaction of methanol is:

$CH_3OH+\frac{3}{2} O_2\rightarrow CO_2+2H_2O$

In such a way, since there is 1:3/2 mole ratio between methanol and oxygen, we can compute the moles of oxygen that are needed to burn 2.56 moles of methanol as shown below:

$n_{O_2}=2.56molCH_3OH*\frac{\frac{3}{2}molO_2}{1molCH_3OH} \\\\n_{O_2}=3.84molO_2$

Best regards!

6 0

3 years ago

The force that keeps the planets in orbit is________.

erma4kov [3.2K]

Answer:

The force that keeps the planets in orbit is gravity.

4 0

3 years ago

Molecules are different from atoms because they

tatyana61 [14]

D) Contain Chemical bonds.

5 0

3 years ago

Read 2 more answers

The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has

Alja [10]

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate law expression is:

$Rate=k[NO]^2[O_2]$

Now we have to determine the number of molecules of $NO\text{ and }O_2$

In mixture 1 : There are 5 $NO$ and 4 $O_2$ molecules.

In mixture 2 : There are 7 $NO$ and 2 $O_2$ molecules.

In mixture 3 : There are 3 $NO$ and 5 $O_2$ molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(5)^2\times (4)$

$Rate=k(100)$

The rate law expression for mixture 2 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(7)^2\times (2)$

$Rate=k(98)$

The rate law expression for mixture 3 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(3)^2\times (5)$

$Rate=k(45)$

Hence, the mixture 1 has the fastest initial rate.

4 0

3 years ago

Helppppp asaapppppp plzzzzzz

Gala2k [10]

Answer:

Alright the very first thing you need to do is balance the equation:

2HCl + Na2CO3 -----> 2NaCl + CO2 + H2O

Now we need to find the limiting reactant by converting the volume to moles of both HCl and Na2CO3.

Volume x Concentration/molarity = moles

0.235L x 0.6 M = 0.141 moles / molar ratio of 2 = 0.0705 moles of HCl

0.094L x 0.75 M = 0.0705 moles /molar ratio of 1 = 0.0705 moles of Na2CO3

Since both of the moles are equal, it means the entire reaction is complete (while the identification of limiting reactant may seem like an unnecessary step, it's quite essential in stoichiometry, so keep an eye out) and there is no excess of any reactant.

Now we know that the product we want to calculate is aqueous so, following the law of conservation of mass, we should add both volumes together to calculate how much volume we could get for NaCl.

0.235 + 0.094 = 0.329L of NaCl

Now we apply the C1V1 = C2V2 equation using the concentration and volume of Na2CO3 because it's molar ratio is one to one to NaCl (You can also use HCL, but you have to divide their moles by 2 for the molar ratio) and the volume we just calculated for NaCl.

(0.75M) x (0.094L) = C2 x (0.329L)

Rearrange equation to solve for C2:

(0.75M) x (0.094L) = C2

(0.329L)

C2 = 0.214 M (Rounded)

When the reaction is finished, the NaCl solution will have a molarity concentration of 0.214 M.

7 0

2 years ago