Silver has metallic bonding.
Silver is a very typical and main metal. The negatively charged electrons distribute themselves throughout the entire piece of metal and form non directional bonds between the positive silver ions, which is metallic bonding, and what silver contributes.
Answer:
T₂ = 19.95°C
Explanation:
From the law of conservation of energy:

where,
mc = mass of copper = 37.2 g
Cc = specific heat of copper = 0.385 J/g.°C
mw = mass of water = 188 g
Cw = specific heat of water = 4.184 J/g.°C
ΔTc = Change in temperature of copper = 99.8°C - T₂
ΔTw = Change in temperature of water = T₂ - 18.5°C
T₂ = Final Temperature at Equilibrium = ?
Therefore,

<u>T₂ = 19.95°C</u>
The initial balanced equation of decomposition is:
2NaN3 —> 2Na + 3N2
Since 1 mole of a gas occupies 24dm^3 at room temperature and pressure:
72dm^3 / 24dm^3 = 3 moles of Nitrogen which required 2 moles of Sodium Azide.
Since moles = mass/Relative molecular mass or molar mass
2 moles x 65~ g/mol = 130g
For the empirical formula:
Element H N
% Mass 2.33% 97.7%
Mass/Mr 2.33 6.97
2.33 7~
Divide by lowest number
1 3
Therefore empirical formula is
H1N3
Answer:

Explanation:
a) Balanced equation
The balanced chemical equation for the titration is

b) pH at start
For simplicity, let's use B as the symbol for NH₃.
The equation for the equilibrium is

(i) Calculate [OH]⁻
We can use an ICE table to do the calculation.
B + H₂O ⇌ BH⁺ + OH⁻
I/mol·L⁻¹: 0.150 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.150 - x x x
![K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.150 - x} = 1.8 \times 10^{-5}](https://tex.z-dn.net/?f=K_%7B%5Ctext%7Bb%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BBH%7D%5E%7B%2B%7D%5D%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BB%5D%7D%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5C%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.150%20-%20x%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D)
Check for negligibility:

(ii) Solve for x
![\dfrac{x^{2}}{0.150} = 1.8 \times 10^{-5}\\\\x^{2} = 0.150 \times 1.8 \times 10^{-5}\\x^{2} = 2.7 \times 10^{-6}\\x = \sqrt{2.7 \times 10^{-6}}\\x = \text{[OH]}^{-} = 1.64 \times 10^{-3} \text{ mol/L}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.150%7D%20%3D%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.150%20%5Ctimes%201.8%20%5Ctimes%2010%5E%7B-5%7D%5C%5Cx%5E%7B2%7D%20%3D%202.7%20%5Ctimes%2010%5E%7B-6%7D%5C%5Cx%20%3D%20%5Csqrt%7B2.7%20%5Ctimes%2010%5E%7B-6%7D%7D%5C%5Cx%20%3D%20%5Ctext%7B%5BOH%5D%7D%5E%7B-%7D%20%3D%201.64%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctext%7B%20mol%2FL%7D)
(iii) Calculate the pH
(c) pH at equivalence point
(i) Calculate the moles of each species

B + HI ⇌ BH⁺ + I⁻
I/mol: 3.00 3.00 0
C/mol: -3.00 -3.00 +3.00
E/mol/: 0 0 3.00
(ii) Calculate the concentration of BH⁺
At the equivalence point we have a solution containing 3.00 mmol of NH₄I
Volume = 20.00 mL + 20.00 mL = 40.00 mL
![\rm [BH^{+}] = \dfrac{\text{3.00 mmol}}{\text{40.00 mL}} = \text{0.0750 mol/L}](https://tex.z-dn.net/?f=%5Crm%20%5BBH%5E%7B%2B%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B3.00%20mmol%7D%7D%7B%5Ctext%7B40.00%20mL%7D%7D%20%3D%20%5Ctext%7B0.0750%20mol%2FL%7D)
(iii) Calculate the concentration of hydronium ion
We can use an ICE table to organize the calculations.
BH⁺+ H₂O ⇌ H₃O⁺ + B
I/mol·L⁻¹: 0.0750 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.0750 - x x x

![\dfrac{x^{2}}{0.0750} = 5.56 \times 10^{-10}\\\\x^{2} = 0.0750 \times 5.56 \times 10^{-10}\\x^{2} = 4.17 \times 10^{-11}\\x = \sqrt{4.17 \times 10^{-11}}\\\rm [H_{3}O^{+}] =x = 6.46 \times 10^{-6}\, mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.0750%7D%20%3D%205.56%20%5Ctimes%2010%5E%7B-10%7D%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.0750%20%5Ctimes%205.56%20%5Ctimes%2010%5E%7B-10%7D%5C%5Cx%5E%7B2%7D%20%3D%204.17%20%5Ctimes%2010%5E%7B-11%7D%5C%5Cx%20%3D%20%5Csqrt%7B4.17%20%5Ctimes%2010%5E%7B-11%7D%7D%5C%5C%5Crm%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3Dx%20%3D%206.46%20%5Ctimes%2010%5E%7B-6%7D%5C%2C%20mol%20%5Ccdot%20L%5E%7B-1%7D)
(iv) Calculate the pH
![\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{6.46 \times 10^{-6}} = \large \boxed{\mathbf{5.19}}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20-%5Clog%7B%5Crm%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%20%3D%20-%5Clog%7B6.46%20%5Ctimes%2010%5E%7B-6%7D%7D%20%3D%20%5Clarge%20%5Cboxed%7B%5Cmathbf%7B5.19%7D%7D)
The titration curve below shows the pH at the beginning and at the equivalence point of the titration.
Answer:
D +405.0kJ mol-¹
Explanation:
Since bond energy is the energy required to break a bond, the energy of dissociation of X₂H₆ = +2775 kJmol⁻¹.
Since there is one X-X bond and six X-H bonds,
Bond energy of one X-X bond + Bond energy of six X-H bonds = energy of dissociation of X₂H₆.
Since bond energy of one X-H bond = 395 kJ mol⁻¹, then
Bond energy of one X-X bond + Bond energy of six X-H bonds = energy of dissociation of X₂H₆
Bond energy of one X-X bond + 6 × one X-H bond = +2775 kJmol⁻¹.
Bond energy of one X-X bond + 6 × 395 kJ mol⁻¹ = +2775 kJmol⁻¹.
Bond energy of one X-X bond + 2370 kJ mol⁻¹ = 2775 kJmol⁻¹
Bond energy of one X-X bond = 2775 kJmol⁻¹ - 2370 kJ mol⁻¹
Bond energy of one X-X bond = +405 kJmol⁻¹