QUËSTIONS :- what are the difference between ionic bond and covalent bond?
IONIC BONDS :- THEY R BONDS WHICH R FORMED DUE TO THE COMPLETE TRANSFER OF ELECTRONS BETWEEN TWO ATOMS
COVALENT BONDS :- THEY R THE BONDS WHICH R FORMED DUE TO INCOMPLETE TRANSFER OF ELECTRONS BETWEEN TWO ATOMS
IONIC BONDS R MORE STRONGER.
Answer:
Solids
Explanation:
Electromagnetic waves travel fastest in empty space. But they are slowest in solids
Answer:
Q = 5350.2 Joules.
Explanation:
<u>Given the following data;</u>
Mass, m = 26. 5g
Initial temperature, T1 = 0°C
Final temperature, T2 = 48.3°C
Specific heat capacity of water, c = 4.18 J/g°C.
*To find the quantity of heat*
Heat capacity is given by the formula;
Where;
- Q represents the heat capacity or quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2 - T1
dt = 48.3 - 0
dt = 48.3°C
Substituting the values into the equation, we have;
Q = 5350.2 Joules.
Therefore, the amount of heat absorbed is 5350.2 Joules.
Answer:
0.525 M CuSO
Explanation:
Molarity (M) is the units mol/L. Let's figure out how many moles of CuSO we have:
35 mL = 35/1000 L = 0.035 L
0.035 L * 1.50 mol/L = 0.0525 mol CuSO
Our new volume is 100 mL, so let's first convert this to L:
100 mL = 100/1000 L = 0.100 L
To find the new molarity, divide the number of moles (0.0525 moles) by the number of liters (0.100 L):
0.0525 mol / 0.100 L = 0.525 M CuSO
Hope this helps!
Answer:
153.3 grams of ZrCl₄ are produced
Explanation:
The equation of the reaction is as follows:
ZrSiO₄ + Cl₂ ----> ZrCl₄ + SiO₂ + O₂
molar mass of ZrSiO₄ = (91 + 32 + 16 * 4) = 187.0 g/mol
molar mass of ZrCl₄ = (91 + 35.5 * 4) = 233.0 g/mol
molar mass of Cl₂ = (35.5 * 2) 71.0 g/mol
From the equation of reaction, 1 mole of ZrSiO₄ reacts with one mole of Cl₂ to produce one mole of ZrCl₄
number of moles of ZrSiO₄ present in 123 g = 123/187 = 0.65 moles
number of moles of Cl₂ present in 85.0 g = 85.0/71.0 = 1.19 moles
therefore, ZrSiO₄ is the limiting reactant
123.0 g of ZrSiO₄ will react with excess Cl₂ to produce 123 * 233/187 grams of ZrCl₄ = 153.3 grams of ZrCl₄
Therefore, 153.3 grams of ZrCl₄ are produced