ammonia is a weak base that will react in water following the equation below. nh3 h2o⟶x oh− what is the chemical formula for the conjugate acid of nh3?

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Answer 1

Ammonia (NH₃) being a weak base, accepts the hydrogen ion from water to form its conjugate acid, ammonium (NH₄⁺).

Ammonia (NH₃) is a weak base that reacts with water (H₂O) to form its conjugate acid and a hydroxide ion (OH⁻) in the process called acid-base reaction. When NH₃ interacts with H₂O, a hydrogen ion (H⁺) from water is transferred to ammonia, resulting in the formation of the conjugate acid of NH₃, which is ammonium (NH₄⁺). At the same time, the hydroxide ion (OH⁻) is produced as a byproduct. The overall balanced equation for this reaction is:

NH₃ (aq) + H₂O (l) ⟶ NH₄⁺ (aq) + OH⁻ (aq)

Here, the chemical formula for the conjugate acid of ammonia (NH₃) is NH₄⁺. It is essential to understand that a conjugate acid is formed when a base accepts a hydrogen ion (H⁺) from the reacting species. In this case, ammonia (NH₃) being a weak base, accepts the hydrogen ion from water to form its conjugate acid, ammonium (NH₄⁺).

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the diffusion coefficient of fe in bcc iron is approximately 3 x 10-11 cm2/s at 900 oc and 1.5 x 10-14 cm2/s at 630oc. the activation energy in cal/mol is approximately

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The activation energy of Fe in BCC iron is approximately 139.06 cal/mol at 900 OC and 199.17 cal/mol at 630 OC.

Given:The diffusion coefficient of Fe in BCC iron is approximately 3 x 10-11 cm2/s at 900 OC and 1.5 x 10-14 cm2/s at 630OCFormula:The Arrhenius equation: k = Ae^(-Q/RT)

Activation Energy, Q = -R ln(k/T)where R is the gas constant, k is the rate constant, T is the absolute temperature, and A is the pre-exponential factor.Calculation:R = 1.987 cal/(mol K)

The activation energy is given byQ=−Rln(kT)At 900 OC: k= 3 x 10-11 cm2/s and T = 1173 KR= 1.987 cal/mol.Kln(kT) = ln(3 x 10^-11 cm²/s × 1173 K) = -69.91 Q = -1.987 cal/(mol K) × (-69.91) Q = 139.06 cal/molAt 630 OC: k = 1.5 × 10-14 cm2/s and T = 903 KR = 1.987 cal/(mol K)ln(kT) = ln(1.5 × 10^-14 cm²/s × 903 K) = -100.32 Q = -1.987 cal/(mol K) × (-100.32) Q = 199.17 cal/mol

Therefore, the activation energy of Fe in BCC iron is approximately 139.06 cal/mol at 900 OC and 199.17 cal/mol at 630 OC.

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