Answer:
Balanced chemical equations only show formulae, not names. A balancing number, written in normal script, multiplies all the atoms in the substance next to it.
Answer:
54
Explanation:
Given symbol of the element:
I⁻
Number of electrons found in an ion with the symbol:
This is a iodine ion:
For an atom of iodine:
Electrons = 53
Protons = 53
Neutrons = 74
An ion of iodine is one that has lost or gained electrons.
For this one, we have a negatively charged ion which implies that the number of electrons is 1 more than that of the protons.
So, number of electrons = 53 + 1 = 54
The number of electrons in this ion is 54
The reaction is:
NH4 (NO3) (s) ⇄ N2O (g) + 2 H2O (g)
This means that 1 mol of NH4 (NO3)s produces 3 moles of gases.
Now find the number of moles in 1.71 kg of NH4 (NO3)
Molar mass = 2*14g/mol + 4 * 1g/mol + 3*16g/mol = 80 g/mol
# moles = mass / molar mass = 1710 g / 80 g/mol = 21.375 mol of NH4(NO3)
We already said that every mol of NH4(NO3) produces 3 moles of gases, then the number of moles of gases produced is 3 * 21.375 = 64.125 mol
Now use the equation for ideal gases to fin the volume
pV = nRT => V = nRT / p = (64.125 mol)(0.082atm*liter / K*mol) * (119 +273)K / (731mmHg *1 atm/760mmHg) =
V = 2143.01 liters
Answer:
Explanation:
To be honest I reallydont know their but hope u get ur answer
Answer:
Volume of water added = 148.4 mL
Explanation:
Given data:
Initial volume = 424 mL
Initial molarity = 0.189 M
Final molarity = 0.140 M
Volume of water added = ?
Solution:
Formula:
M₁V₁ = M₂V₂
0.189 M×424 mL = 0.140 M×V₂
V₂ = 0.189 M×424 mL /0.140 M
V₂ = 80.136 M.mL / 0.140 M
V₂ = 572.4 mL
Final volume of solution is 572.4 mL.
Volume of water added = Final volume - initial volume
Volume of water added = 572.4 mL - 424 mL
Volume of water added = 148.4 mL