Here we have to choose the right option which tells the moles of CaCl₂ will react with 6.2 moles of AgNO₃ in the reaction
2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂
6.2 moles of silver nitrate (AgNO₃) will react with B. 3.1 moles of calcium chloride (CaCl₂).
From the reaction: 2AgNO₃ + CaCl₂→ 2AgCl + Ca(NO₃)₂
Thus 2 moles of AgNO₃ reacts with 1 mole of CaCl₂
Henceforth, 6.2 moles of AgNO₃ reacts with 
= 3.1 moles of CaCl₂.
1 mole of CaCl₂ reacts with 2 moles of AgNO₃. Thus-
A. 2.2 moles of CaCl₂ will react with 2.2×2 = 4.4 moles of AgNO₃.
C. 6.2 moles of CaCl₂ will reacts with 6.2×2 = 12.4 moles of AgNO₃.
D. 12.4 moles of CaCl₂ will reacts with 12.4 × 2 = 24.8 moles of AgNO₃
Thus the right answer is 6.2 moles of AgNO₃ will react with 3.1 moles of CaCl₂.
Answer:
that are formed from hybridized orbitals
Explanation:
The chemical concept of resonance is when a change in the position of the electrons occurs, without changing the position of the atoms.
The structure obtained in the resonance will not be any of the previous ones, but a hybrid of resonance between those structures.
There’s no question for me to answer ?
Answer: a) 
b) 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
a) Mass of Ba= 66.06 g
Mass of Cl = 34.0 g
Step 1 : convert given masses into moles.
Moles of Ba =
Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Ba = 
For O =
The ratio of Ba: Cl= 1:2
Hence the empirical formula is
b) Mass of Bi= 80.38 g
Mass of O= 18.46 g
Mass of H = 1.16 g
Step 1 : convert given masses into moles.
Moles of Bi =
Moles of O= 
Moles of H=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Bi= 
For O =
For H=
The ratio of Bi: O: H= 1:3: 3
Hence the empirical formula is
Answer:
Explanation:
HCl + NaOH ⟶ NaCl + H₂O
There are two energy flows in this reaction.
Heat of reaction + heat to warm water = 0
q₁ + q₂ = 0
q₁ + mCΔT = 0
Data:
m(HCl) = 50 g
m(NaOH) = 50 g
T₁ = 22 °C
T₂ = 28.87 °C
C = 4.18 J·°C⁻¹g⁻¹
Calculations:
m = 50 + 50 = 100 g
ΔT = 28.87 – 22 = 6.9 °C
q₂ = 100 × 4.18 × 6.9 = 2900 J
q₁ + 2900 = 0
q₁ = -2900 J
The negative sign tells us that the reaction produced heat.
The reaction produced 
.
