The ease which an acid or base dissociates in a solution refers to

larisa86 [58]

The balanced reaction is:

4NH3 + 3O2 --> 2N2 + 6H2O

<span>We are given the amount of reactants to be used for the reaction. This will be the starting point of our calculation.</span>

83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2

2.81 moles of NH3

From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.

8 0

3 years ago

The atomic number of nitrogen is 7. How many protons, neutrons, and electrons make up an atom of Nitrogen-15?

earnstyle [38]

<h2><u>Answer:</u></h2>

A nonpartisan iota of Nitrogen has a mass of 18. There are 7 protons in the core of this iota. What number of neutrons, complete electrons, and valence electrons are available

Nitrogen 15 has a nuclear mass of 15. The mass number is # protons in addition to # of neutrons, so for N-15 mass is 15 and the protons are dependably 7 so there must be 15-7=8 neutrons. N-15 has 7 electrons since it has 7 protons and p = e.

6 0

3 years ago

Match the academic requirements with the careers . Cosmetologist

inessss [21]

Answer:

Technical program - cosmetologist

5 0

4 years ago

At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO

erastovalidia [21]

Answer : The concentration of $NO$ at equilibrium is 0.9332 M

Solution : Given,

Concentration of $N_2$ and $O_2$ at equilibrium = 0.200 M

Concentration of $N_2$ and $O_2$ at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}$

$K_c=6.25$

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially 0.200 0.200 0

.800

At equilibrium (0.200-x) (0.200-x) (0.800+2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}$

By solving the term x, we get

$x=2.6\text{ and }-0.0666$

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of $NO$ at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of $NO$ at equilibrium is 0.9332 M

5 0

4 years ago

State one alternative to burning fossil fuels in order to produce electricity? Please answer

Oksanka [162]

Answer:

Some well-known alternative fuels include bio-diesel, bio-alcohol (methanol, ethanol, butane), refuse-derived fuel, chemically stored electricity (batteries and fuel cells), hydrogen, non-fossil methane, non-fossil natural gas, vegetable oil, propane and other biomass sources

8 0

3 years ago