Answer:
CaF2 will not precipitate
Explanation:
Given
Volume of Ca(NO3)2 ml
Molar concentration of Ca(NO3)2
Volume of NaF ml
Molar concentration of NaF
Ksp for CaF2
CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2
Moles of calcium ion
Moles of F- ion
Q is lesser than Ksp value of CaF2. Hence it will not precipitate
Answer:
radiation, conduction, convection, conduction
Answer:
0.482 ×10²³ molecules
Explanation:
Given data:
Volume of gas = 2.5 L
Temperature of gas = 50°C (50+273 = 323 k)
Pressure of gas = 650 mmHg (650/760 =0.86 atm)
Molecules of N₂= ?
Solution:
PV= nRT
n = PV/RT
n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k
n = 2.15 atm. L /26.52 atm. mol⁻¹.L
n = 0.08 mol
Number of moles of N₂ are 0.08 mol.
Number of molecules:
one mole = 6.022 ×10²³ molecules
0.08×6.022 ×10²³ = 0.482 ×10²³ molecules
<h2>
Hey There!</h2><h2>
_____________________________________</h2><h2>
Answer:</h2><h2>
----------------------------------------------------------------------</h2><h2>
ELECTROPLATING </h2>
Electroplating the plating one metal on to the another metal, It is mostly used for preventing corrosion by using copper or chromium or decorate the object by using gold or silver plating.
__________________________________________________________
<h3>Principle:</h3>
When electricity is passed the thin layer of metal is deposited on another metal and water molecule given out as a By-Product, Thus this process works on the principle of Hydrolysis.
<h2>_____________________________________</h2><h2>QUESTION:</h2>
A) Electrolysis
B) Chromium prevents corrosion and gives the fine shining touch to the objects.
C) The metal which is deposited to the object i.e. spoon will be connected to the positive electrode of a battery, Thus it is anode. The spoon at which electroplating is need to be done is connected to the negative electrode, thus the Spoon is cathode.
<h2>
_____________________________________</h2><h2>
Best Regards,</h2><h2>
'Borz'</h2>
Answer:
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced
Now use the ideal gas law to find the volume V.
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm
T = 21 C + 273 = 294 K
PV = nRT
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L