Answer:
Mass = 76.176 g
Explanation:
Given data:
Mass of lead(II) chloride produced = 62.9 g
Mass of lead(II) nitrate used = ?
Solution:
Chemical equation:
Pb(NO₃)₂ + 2HCl → PbCl₂ + 2HNO₃
Number of moles of lead(II) chloride:
Number of moles = mass/molar mass
Number of moles = 62.9 g/ 278.1 g/mol
Number of moles = 0.23 mol
Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:
PbCl₂ : Pb(NO₃)₂
1 : 1
0.23 : 0.23
Mass of Pb(NO₃)₂:
Mass = number of moles × molar mass
Mass = 0.23 mol × 331.2 g/mol
Mass = 76.176 g
Answer:
The ΔH of the reaction is + 12.45 KJ/mol
Explanation:
Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)
heat capacity of water = 4.18 Jk-1 Mol-1
Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)
Molar mass of NaHCO3 = 84 g/mol
Mole of NaHCO3 = 14.5 / 84 = 0.173 mol
Step 1 : Calculate the heat energy (Q) lost by the water.
Q = M x C x ΔT
Q = -100 x 4.18 x (-5.14)
Q = 2148.5 joules
Q = 2.1485 K J
Step 2: Calculating the ΔH of the reaction?
ΔH = Q / number of moles of NaHCO3
ΔH = 2.1485 / 0.173
ΔH = 12.42 KJ/mol
Given :
Number of atoms of an element, n = 2.072 × 10⁴³ atoms.
To Find :
Number of moles of that element in given number of atoms.
Solution :
We know, 1 mole of any element contains 6.022 × 10²³ atoms.
So, number of moles in given number of atoms are :

Hence, this is the required solution.
Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)


Given s = 7.4×10⁻³ M
So, Ksp is:


Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)

Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>
matter can take be one of the three states which are; (Solid, Liquid, Gas)
Think about water, it can be a Solid (Ice) a Liquid (not frozen obviously) and a Gas (when it evaporates)