Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
1. Assemble all the data in one place, with molar masses above the formulas and other information below them.
Mᵣ: 58.44
NaCl + AgNO₃ ⟶ NaNO₃ + AgCl
m/g: 0.245
V/mL: 50.
c/mmol·mL⁻¹: 0.0180
2. Calculate the moles of each reactant
3. Identify the limiting reactant
Calculate the moles of AgCl we can obtain from each reactant.
From NaCl:
The molar ratio of NaCl to AgCl is 1:1.
From AgNO₃:
The molar ratio of AgNO₃ to AgCl is 1:1.
AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.
4. Calculate the moles of excess reactant
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
I/mmol: 0.900 4.192 0
C/mmol: -0.900 -0.900 +0.900
E/mmol: 0 3.292 0.900
So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.
5. Calculate the concentration of Cl⁻
Answer: One mol of NaCl (6.02 x1023 formulas) has a mass of 58.44 g.
To go from grams to moles, divide the grams by the molar mass. 600 g58.443 g/mol = 10.27 mol of NaCl.
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Explanation:
Answer:
:)
Explanation:
Recycling paper helps to reduce greenhouse gas emissions that can contribute to climate change. It takes 70% less energy and water to recycle paper than to create new paper from trees. Manufacturing with recovered paper cuts down on pollution that contributes to smog (and ill health).
Answer: v = 2π2 Kme2 Z / nh
Explanation:
The formula for velocity of an electron in the nth orbit is given as,
v = 2π2 Kme2 Z / nh
v = velocity
K = 1/(4πε0)
m= mass of an electron
e = Charge on an electron
Z= atomic number
h= Planck’s constant
n is a positive integer.