1.Which statement best describes an oxidation-reduction reaction?(1 point)

The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.

harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

Solution:- The balanced equation for the combustion of acetylene is:

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)$

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})$

= $70molCO_2$

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})$

= $87.5molO_2$

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

$84.0molO_2(\frac{4molO_2}{5molO_2})$

= $67.2molCO_2$

7 0

4 years ago

Read 2 more answers

A student placed equal volumes of honey and of water in two, identical, open dishes, and left them at room temperature for 8 hou

Ahat [919]

Answer: Honey has a much lower vapor pressure than pure water has. So, pure water evaporates at a much higher rate.

Explanation:

8 0

3 years ago

Read 2 more answers

. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

3 0

3 years ago

What is the total number of calories of heat energy absorbed when 10 grams of water is vaporized at its normal boiling point

Darya [45]

<u>Answer:</u> The amount of energy absorbed by water is 5390 Calories

<u>Explanation:</u>

To calculate the amount of heat absorbed at normal boiling point, we use the equation:

$q=m\times L_{vap}$

where,

q = amount of heat absorbed = ?

m = mass of water = 10 grams

$L_{vap}$ = latent heat of vaporization = 539 Cal/g

Putting values in above equation, we get:

$q=10g\times 539Cal/g=5390Cal$

Hence, the amount of energy absorbed by water is 5390 Calories

6 0

3 years ago

What volume will 1.27 moles of helium gas occupy at 80.00 °C and 1.00 atm?

White raven [17]

Answer:

36.8 L

Explanation:

We'll begin by converting 80 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 80 °C

T(K) = 80 + 273

T(K) = 353 K

Finally, we shall determine the volume occupied by the helium gas. This can be obtained as follow:

Number of mole (n) = 1.27 moles

Temperature (T) = 353 K

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =?

PV = nRT

1 × V = 1.27 × 0.0821 × 353

V = 36.8 L

Thus, the volume occupied by the helium gas is 36.8 L

5 0

3 years ago