Missing question: <span>Assume a density of 10.35 g/cm3 for Ag, A(Ag) = 107.87 g/mol.
N(Ag) = Na </span>· d(Ag) ÷ A(Ag).
N(Ag) = 6,023·10²³ atoms/mol · 10,35 g/cm³ · 10⁶ cm³/m³ ÷ 107,87 g/mol.
N(Ag) = 5,78·10²⁸ atoms/mol.
Nv = 5,78·10²⁸ atoms/mol · 5·10⁻⁵.
Nv = 2,89·10²².
Answer:
b: LEADS
Explanation:
i guessed and got it right ahh
<h3>
Answer:</h3>
28.96 kJ/°C
<h3>
Explanation:</h3>
We are given;
- Enthalpy change (ΔH) = −3226.7 kJ/mol
- The reaction is exothermic since the heat change is negative;
- Mass of benzoic acid = 3.1007 g
- Temperature change (21.84°C to 24.67°C) = 2.83°C
We are required to find the heat capacity of benzoic acid;
<h3>Step 1: Moles of benzoic acid </h3>
Moles = Mass ÷ molar mass
Molar mass of benzoic = 122.12 g/mol
Therefore;
Moles = 3.1007 g ÷ 122.12 g/mol
= 0.0254 moles
<h3>Step 2: Determine the specific heat capacity </h3>
Heat change for 1 mole = 3226.7 kJ
Moles of Benzoic acid = 0.0254 moles
But;
Specific heat capacity × ΔT = Moles × Heat change
cΔT = nΔH
Therefore;
Specific heat capacity,c = nΔH ÷ ΔT
= (3226.7 kJ × 0.0254 moles) ÷ 2.83°C
= 28.96 kJ/°C
Therefore, the specific heat capacity of benzoic acid is 28.96 kJ/°C
6.75 moles of H₂
Explanation:
We have the following balanced chemical reaction:
Mg + 2 HCl → MgCl₂ + H₂
From the chemical reaction we deduce that if 1 mole of Mg is reating with 2 moles of HCl then 8.30 moles of Mg is reaction with 16.60 moles of HCl, quantity which is over our available 13.5 moles of HCl. The limiting reactant is HCl.
Knowing this we devise the following reasoning:
if 2 moles of HCl produces 1 mole of H₂
then 13.5 moles of HCl produces X moles of H₂
X = (13.5 × 1) / 2 = 6.75 moles of H₂
Learn more about:
balancing chemical equations
brainly.com/question/13898428
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To create the Lewis structure we need to take into account the octet rule: atoms tend to gain, lose or share electrons to complete their valence shell with 8 electrons.
C belongs to Group 4A in the periodic table so it has 4 valence electrons. It needs to share 4 pairs of electrons to complete the octet.
F belongs to Group 7A in the periodic table so it has 7 valence electrons. Each F needs to share 1 pair of electrons to complete the octet.
As a consequence, in CF₄, C will form a single bond with each F and all the octets will be complete.