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JulijaS [17]
3 years ago
12

Write the balanced chemical equation for the Haber-Bosch process, that is, the combination of nitrogen and hydrogen to form ammo

nia, NH3. Phase symbols are optional.
Chemistry
2 answers:
RoseWind [281]3 years ago
6 0
The balanced chemical equation for the Haber-Bosch process is N₂(g) + 3H₂(g) → 2NH₃(g). The Haber-Bosch process played a significant role in boosting agriculture back in the day. It paved the way for the industrial production of ammonia which is used in the manufacture of fertilizers. The process involves reacting atmospheric N₂ with H₂ using a metal catalyst under high temperature and pressure.    
irina [24]3 years ago
3 0

Answer: N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

Explanation: According to the law of conservation of mass, mass can neither be created nor be destroyed. Mass remains conserved. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Haber's process used for manufacturing of ammonia in terms of balanced chemical equation is:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

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N2 + 3H2 + 2NH3 .
HACTEHA [7]

Answer:

7

Explanation:

7 0
3 years ago
Two atoms of the same element
kotykmax [81]

Answer:

Isotope

Explanation:

5 0
2 years ago
In the following reaction, which compounds are considered based? HNO2+H2O <—>H3O + NO2
ehidna [41]

Answer:

non of them are considered base hno3 is acidic ,H2O is neutral ,h30 is acidic likewise n02

7 0
3 years ago
A 55-mL solution of H2SO4 is completely neutralized by 46 mL of 1.0M NaOH. What is the concentration of the H2SO4∆H2SO4(aq) +2Na
slega [8]
H₂SO₄ + 2NaOH = Na₂SO₄ + 2H₂O

v(NaOH)=46 ml=0.046 l
c(NaOH)=1.0 mol/l
v(H₂SO₄)=55 ml=0.055 l

n(NaOH)=v(NaOH)*c(NaOH)

n(H₂SO₄)=0.5n(NaOH)

c(H₂SO₄)=n(H₂SO₄)/v(H₂SO₄)=0.5*v(NaOH)*c(NaOH)/v(H₂SO₄)

c(H₂SO₄)=0.5*0.046*1.0/0.055=0.418 mol/l

The concentration of the H₂SO₄ is 0.418M.

4 0
3 years ago
What volume of a 0.3300M solution of sodium hydroxide would br ruired to titrate 15.00 mL of 0.1500 M oalic Acid?
Ganezh [65]

Answer:

3.41 mL

Explanation:

At equivalence point  from the reaction given,

Moles of Oxalic\ Acid = 2 × Moles of NaOH

Considering

Molarity_{Oxalic\ Acid}\times Volume_{Oxalic\ Acid}=2\times Molarity_{NaOH}\times Volume_{NaOH}

Given  that:

Molarity_{NaOH}=0.3300\ M

Volume_{NaOH}=?

Volume_{Oxalic\ Acid}=15.00\ mL

Molarity_{Oxalic\ Acid}=0.1500\ M

So,  

Molarity_{Oxalic\ Acid}\times Volume_{Oxalic\ Acid}=2\times Molarity_{NaOH}\times Volume_{NaOH}

2\times 0.3300\times Volume_{NaOH}=0.1500\times 15.00

Volume_{NaOH}=3.41\ mL

4 0
2 years ago
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