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Alecsey [184]
3 years ago
11

Balance the equation and determine how many liter of nitrogen are required to react with 7.3 L of hydrogen to produce ammonia (n

itrogen trihydride)?
Chemistry
1 answer:
Musya8 [376]3 years ago
3 0

Answer:

4.867 L of ammonia

Explanation:

Using Haber's process to form ammonia using Nitrogen and hydrogen, the equation is :

N₂ + 3H₂ → 2NH₃

Here, 3 moles of hydrogen gas gives 2 moles of ammonia.

1 mole of any substance occupies 22.4L at STP

So, 3 x 22.4L of hydrogen gives 2 x 22.4 L of ammonia

Then 7.3 L of hydrogen will give:

=\frac{7.3 \times 2 \times22.4 }{3 \times22.4}

=\frac{7.3 \times 2}{3}

= 4.867 L of ammonia

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the water underwent a physical change. This is because the water simply changed states, and could become liquid again, but it will remain H20 throughout the time

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2 years ago
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator
Eduardwww [97]

Answer:

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

Bromocresol green, color change from pH = 4.0 to  5.6

Explanation:

The equation for the reaction is :

C_2H_5NH_2_(_a_q_)     +     H^+_(_a_q_)   ---      C_2H_5NH_{3(aq)}^+

concentration of C_2H_5NH_{2(aq) = 10%

10 g of C_2H_5NH_{2(aq) in 100 ml solution

molar mass = 45.08 g/mol

number of moles = 10 / 45.08

= 0.222 mol

Molarity of C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL

= 2.22 M

number of moles of C_2H_5NH_{2(aq) in 20 mL can be determined as:

= 20 mL \times  2.22 M= 44*10^{-3} mole

Concentration of C_2H_5NH_2(aq) = \frac{44*10^{-3}*1000}{20}

= 2.22 M

Similarly, The pKa Value of C_2H_5NH_{2(aq) is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

the pH value at equivalence point is,

pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]

pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]

pH = 5.21

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

8 0
3 years ago
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