All categories

1 year ago

Below is the thermochemical equation for the combustion of octane.

Chemistry

1 answer:

Gnom [1K]1 year ago

8 0

5mol produces -10.941KJ

4 mol let produces x KJ

Solve through proportion

5/-10.941=4/x
5x=-43.764
x=-43.764/5
x=8.756 KJ

You might be interested in

Protects the eyes from flying objects or chemical splashes

Bogdan [553]

Answer:

Goggles

Explanation:

5 0

3 years ago

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe

Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

$V_s=V_o-V_i=V_i\\V_o=2*V_i\\$

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

$r_o=r_i+e$

The first equation becomes

$\frac{4 \pi}{3}*r_o^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\\frac{4 \pi}{3}*(r_i+e)^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\(r_i+e)^{3} = 2 *r_i^{3} \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0$

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

8 0

2 years ago

Elements in the same period have the same…

ch4aika [34]

Answer:

Elements in the same period have the same (number of electron shells)

8 0

2 years ago

Where does fusion regularly occur and what kind of energy is produced?

Pani-rosa [81]

The answer is <span>B. Fusion occurs inside the sun. It creates light and heart that sustain life on earth.

The Sun creates its own energy through fusion. This also happens in larger stars as well.

Have a good day! =)</span>

5 0

2 years ago

Read 2 more answers

A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate. 2KClO3 mc019-1.jpg 2KCI + 3O2 What is the perc

Anon25 [30]

The balanced reaction is 2KClO3 --> 2KCl + 3O2
We first divide the 400.0 g KClO3 by the molar mass of 122.55 g/mol to get 3.26 mol KClO3. Next, we use the coefficients: 3.26 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 4.896 mol O2. Multiplying this by the molar mass of 32 g/mol gives 156.67 g O2.
Percent yield = 115.0 g / 156.67 g = 0.734 = 73.40%

3 0

3 years ago