300 °C would be found in which temperature range?

Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3 (g). At equilibrium, th

artcher [175]

Answer:

The <u>equilibrium constant</u> is:

$k_c=0.0030M^{-2}$

Explanation:

The correct equation is:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.

The equation for the equilibrium constant is:

$k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}$

Substituting:

$k_c=\dfrac{(0.105M)^2}{(1.1M)\cdot (1.50M)^3}$

$k_c=0.0030M^{-2}$

6 0

3 years ago

A family pool holds 10,000 gallons of water. How many cubic centimeters is this?

Ierofanga [76]

Answer:

3.785412e+7cm³

Explanation:

Have a nice day.

3 0

2 years ago

How many moles are there in 7.5 L Of H2

True [87]

So multiply number of moles x number of atoms/mole = 1.8066 x 10^24 atoms of H2. One mole of any gas at STP has a volume of 22.4 L. So first determine the number of moles of gas you have.
for example do 7
$7.5 \div 22.4$
that 's what I think

3 0

3 years ago

____ 26. what is the weight percentage of nitrogen in urea, cn2h4o

Paraphin [41]

Weight percentage of nitrogen can be calculated using the following rule:
weight percentage of nitrogen = (weight of nitrogen / weight of urea) x 100

From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of nitrogen = 14 grams
molecular mass of hydrogen = 1 grams
molecular mass of oxygen = 16 grams

therefore:
mass of nitrogen in urea = 2(14) = 28 grams
mass of urea = 12 + 2(14) + 4(1) + 16 = 60 grams

Substitute with the masses in the equation to get the percentage:
weight percentage of nitrogen = (28/60) x 100 = 46.667%

7 0

3 years ago

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0

4 years ago