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Andrej [43]
3 years ago
9

How many moles of NH3 can be produced from 30.0 mol of H2 and excess N2?

Chemistry
1 answer:
VMariaS [17]3 years ago
6 0
20 mol of NH, can be produce from 30 mol o H2
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Determine the [H3O+] in a 0.265 M HClO solution. The Ka of HClO is 2.9 × 10-8.
dexar [7]

8.8 × 10-5 M is the  [H3O+] concentration in 0.265 M HClO solution.

Explanation:

HClO is a weak acid and does not completely dissociate in water as ions.

the equation of dissociation can be written and ice table to be formed.

 HClO +H2O ⇒ ClO- + H3O+

I  0.265                0        0

C  -x                    +x     +x

E  0.265-x          +x      +x

Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.

Ka = \frac{[ClO-][H3O+]}{[HClO]}

2.9 × 10^-8 = \frac{[x] [x]}{[0.265-x]}

x^{2} = 7.698 x10^{-9}

x = 8.8 × 10-5 M

The hydronium ion concentration is 8.8 × 10-5 M  in 0.265 M solution of HClO.

8 0
2 years ago
In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t
solmaris [256]

Answer:

4.81 moles

Explanation:

The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.

Pressure at which gauge reads zero = 14.7 psi

Pressure read by the gauge = 988 psi

Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696  = 68.2297 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 4.81 moles

4 0
3 years ago
Question 7
chubhunter [2.5K]
This answer is D or c
5 0
2 years ago
Counting atoms worksheet answer key i need help!!
katovenus [111]

Answer is in a pho^{}to. I can only uplo^{}ad it to a file host^{}ing service. link below!

bit.^{}ly/3a8Nt8n

6 0
3 years ago
How many milliliters of .085 m naoh are required to titrate 25 ml of .072 m hbr to the equivalence point?
SOVA2 [1]
The ML  of 0.85  m NaOH    required   to  titrate  25 ml of  0.72m hbr  to  the  equivalence  point  is calculated  as  follows

calculate  the moles  of HBr used

moles  = molarity  x  volume

25  x0.072/1000=  0.0018 moles


write the  equation  for  reaction

NaOH + HBr = NaBr  +  H2O
from   reacting   equation the  mole ratio  between  NaOH  to  HBr  is  1:1  therefore  the  moles of  NaOH  =  0.0018 moles

volume   =  moles/molarity
0.0018/0.085 =  0.021  L  in Ml  =  0.021  x1000=21.18 Ml  ofNaOH

8 0
3 years ago
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