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bagirrra123 [75]
2 years ago
15

A vessel contains 18.06*10^23 molecules of ammonia gas at 25°c. Calculate the number of hydrogen atoms present in the ammonia ga

s.
Chemistry
2 answers:
velikii [3]2 years ago
8 0

Answer:

Explanation:

<h2>Before you use log you have to multiply the answer of 8.66 mol by 6.02⋅1023 by 3 because the answer of 8.66 mol by 6.02⋅1023 gives you the number of ammonia molecules and in each molecules there are 3 hydrogen atoms present. </h2><h2> </h2><h2>So it is: </h2><h2> </h2><h2>8.66⋅6.02⋅1023=5.21332⋅1024 </h2><h2>5.21332⋅1024⋅3=1.563996⋅1025 </h2><h2>log(1.563996⋅1025)=25.19 </h2>

Elena L [17]2 years ago
4 0

Answer:There are 5.418×10^24 hydrogen atoms present in the ammonia.

Explanation:

Ammonia (NH3) is a molecule that exists as a gas at Standard temperature (298 K). It contains atoms of Nitrogen and Hydrogen in atomic ratio of 1:3. We can represent it in equation as:

N2 + 3H2 <------> 2NH3

One molecule of NH3 contains one atom of Nitrogen (N) and three atoms of Hydrogen (H). That is 3 atoms of Hydrogen are present in every 1 molecule of NH3. We can as well say that 1 atom of Hydrogen is present in 1/3 molecule of NH3.

Number of NH3 molecules = 1/3 × number of hydrogen atoms

Number of NH3 molecules = 18.06 × 10^23 molecules at 25°C

18.06× 10^23 molecules = 1/3 × number of hydrogen atoms

Cross multiply,

Number of hydrogen atoms= 3× 18.06×10^23 molecules

Number of hydrogen atoms= 5.418×10^24 atoms

Therefore, there are 5.418×10^24 hydrogen atoms present in the ammonia.

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From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2
sashaice [31]

Answer:

From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

(Ans)

ΔHf° of CaC2 = -59.0 kJ/mol

Explanation:

CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ

ΔHrxn = −127.2kJ

ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);

ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn

Where

ΔHf°(C2H2) = 227.4 kJ/mol

ΔHf°(H2O) = -285.8 kJ/mol and

ΔHf°(Ca(OH)2) = -985.2 kJ/mol

ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol

ΔHf°(CaC2) = -59.0 kJ/mol

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     0.89 = (0.5)(t / 5730 years)

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<em>Answer: 963 years</em>
8 0
2 years ago
Consider the following balanced thermochemical equation for a reaction sometimes used for H2S production:
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Answer:

d. Heat is released from the reaction

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2 years ago
What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)
blsea [12.9K]
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, (C_{2}H_{5})_{3}N, the reaction will be
(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}

 and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value  = 5.3 x 10^-4

And kb =  \frac{( [( C_{2}H_{5})_{3}NH^{+} ]*  OH^{-} )}{[( C_{2}H_{5})_{3}N]}

 thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

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2 years ago
A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimete
N76 [4]

Answer:

The mass of water = 219.1 grams

Explanation:

Step 1: Data given

Mass of aluminium = 32.5 grams

specific heat capacity aluminium = 0.921 J/g°C

Temperature = 82.4 °C

Temperature of water = 22.3 °C

The final temperature = 24.2 °C

Step 2: Calculate the mass of water

Heat lost = heat gained

Qlost = -Qgained

Qaluminium = -Qwater

Q = m*c*ΔT

m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)

⇒with m(aluminium) = the mass of aluminium = 32.5 grams

⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C

⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C =  -58.2 °C

⇒with m(water) = the mass of water = TO BE DETERMINED

⇒with c(water) = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C

32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9

-1742.1 = -7.95m

m = 219.1 grams

The mass of water = 219.1 grams

8 0
2 years ago
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