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yanalaym [24]
2 years ago
5

Given the balanced equation representing a reaction: 4nh3 (g) + 5o2 (g) → 4no(g) + 6h2o(g) what is the number of moles of h2o(g)

formed when 2.0 moles of nh3 (g) react completely?
Chemistry
2 answers:
alexdok [17]2 years ago
6 0
To solve, you find the correct mole ratio from the given equation. The questions asks for how many moles of H2O from 2.0 moles of NH3(ammonia), so the mole ration is going to be 6:4.
6 moles of  H2O and 4 moles of NH3.

<u>Tip:</u>
We multiply the given with what we want divided by what we want to get rid of.

So, 2.0mol NH3 x 6mol H2O /4mol NH3.


The answer is  3 moles of H2O or 3mol H2O
pav-90 [236]2 years ago
4 0

Answer : The number of moles of H_2O formed is, 3 moles

Solution : Given,

Moles of NH_3 = 2.0 moles

Now we have to calculate the moles of H_2O

The given balanced chemical reaction is,

4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)

From the balanced reaction, we conclude that

As, 4 moles of NH_3 react to give 6 moles of H_2O

So, 2.0 moles of NH_3 react to give \frac{6}{4}\times 2.0=3 moles of H_2O

Therefore, the number of moles of H_2O formed is, 3 moles

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Answer:

A

Explanation:

The mass number (represented by the letter A) is defined as the total number of protons and neutrons in an atom. Consider the table below, which shows data from the first six elements of the periodic table. Consider the element helium. Its atomic number is 2, so it has two protons in its nucleus.

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How do you determine valence electrons of a metal? A non Metal?
Harlamova29_29 [7]

Answer:

Check the electronic configuration of elements.

Explanation:

▪Valence electrons are the elwctrons present in the outermost shell of any element.

For example,

Electronic Configuration of Sodium = 2,8,1

Here , Sodium has 1 valence electrons.

▪Valency of an element is the total no. of electrons to be gained/losed in order to achieve duplet/octate state.

For example,

Electronic configuration of Sodium = 2,8,1

Sodium can achieve octate state either by losing 1 electron or gaining 7 electrons. But losing 1 electron is eay than gaining 7 electrons. So Valency of Sodium = +1

☆Metals have 1 or 2 or 3 valence electrons.

☆Non metals have 4 or 5 or 6 or 7 valence electrons.

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).
Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas         Mole Fraction      kH mol/(L*atm)

           N_2                         7.81\times 10^{-1}         6.70\times 10^{-4}

           O_2                         2.10\times 10^{-1}        1.30\times 10^{-3}

           Ar                          9.34\times 10^{-3}        1.40\times 10^{-3}

          CO_2                        3.33\times 10^{-4}        3.50\times 10^{-2}

          CH_4                       2.00\times 10^{-6}         1.40\times 10^{-3}

          H_2                          5.00\times 10^{-7}         7.80\times 10^{-4}

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}

where,

p_A = partial pressure of hydrogen gas = ?

p_T = total pressure = 0.380 atm

\chi_A = mole fraction of hydrogen gas = 5.00\times 10^{-7}

Putting values in above equation, we get:

p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{H_2}=K_H\times p_{H_2}

where,

K_H = Henry's constant = 7.80\times 10^{-4}mol/L.atm

p_{H_2} = partial pressure of hydrogen gas = 1.9\times 10^{-7}atm

Putting values in above equation, we get:

C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is 1.48\times 10^{-10}M

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