solution:
1000 = m*2400*(78-22) + m*8.79*10^5
1000= 134400m + 879000m
1000= 1030200m
m = 1000/1013400
m= 1013.4 grams
the final answer is 0.9706 grams
Answer:
<u>One lone-Pair is present in Ammonia</u>
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Explanation:
The number of valence electron in N = 5
The number of Valence electron in H = 1
The formula of ammonia = NH3
Total valence electron in ammonia molecule = 5 +3(1) = 5+3 = 8
The lewis structure suggest that :
Nitrogen completes its octet by sharing the electron pair with 3 hydrogen atoms.
3 electron of Nitrogen are involved in sharing with Hydrogen
So,<u><em> remaining two electron are left non-bonded</em></u> . Hence they exist as lone- pair
So, there is only 1 lone pair in the ammonia molecule .
The shape of NH3 is bent according to VSEPR theory . This is so because the presence of 1 lone pair causes more repulsion and occupy more space.
Thus the lone pair is changing the shape of the ammonia molecule . It also increase the dipole moment of the molecule , which gives polarity to it.
1) Silicon dioxide formula: SiO2 ....... 2 is a subscript for the O atom
2) From the formula you have 1 molecula of SiO2 contains 1 atom of SiO2
3) Then, 0.100 mol of SiO2 contains 0.1 mol of Si
4) Multiply by Avogadro's number: 0.100mol * 6.022*10^23 atoms/mol= 6.02*10^22 atoms
Answer: 6.02*10^22 atoms
Answer:
The balanced chemical equation: NH₃ + 2 HF → NH₄⁺ + HF₂⁻
Explanation:
According to the Brønsted–Lowry acid–base theory, the acid- base reaction is a type of chemical reaction between the acid and base to give a conjugate acid and a conjugate base.
In this reaction, a Brønsted–Lowry acid loses a proton to form a conjugate base. Whereas, a Brønsted–Lowry base accepts a proton to form a conjugate acid.
Acid + Base ⇌ Conjugate Base + Conjugate Acid
The acid dissociation constant (Kₐ) <em>signifies the acidic strength of a chemical species.</em>
∵ pKₐ = - log Kₐ
Thus for a strong acid, Kₐ value is large and pKₐ value is small.
pKₐ (HF) = 3.2 → strong acid
pKₐ (NH₃) = 38 → weak acid
<u>The chemical reaction involved in the dissolution process:</u>
NH₃ + 2 HF → NH₄⁺ + HF₂⁻
In this acid-base reaction, the acid HF reacts with NH₃ base to give the conjugate base HF₂⁻ and conjugate acid NH₄⁺.
<u>HF (acid) donates a proton to form the conjugate base, HF₂⁻ ion. NH₃ (base) accepts a proton to form the conjugate acid. </u>
