Answer:
1B +4St+1Y+3lc——-> BSt4Ylc3
Explanation:
I only know the answer for the first question.
Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT. Substituting a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
The experimental mole ratio of silver chloride to barium chloride is calculated as below
fin the mole of each compound
mole= mass/molar mass
moles of AgCl = 14.5g/142.5 g/mol = 0.102 moles of AgCl
moles of BaCl2 = 10.2 g/208 g/mol = 0.049 moles of BaCl2
find the mole ratio by dividing each mole with the smallest mole(0.049)
AgCl= 0.102/0.049 =2
BaCl2 = 0.049/0.049 =1
therefore the mole ratio AgCl to BaCl2 is 2 :1
D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
Here are some examples of chemical properties:
Reactivity with other chemicals.
Toxicity.
Coordination number.
Flammability.
Enthalpy of formation.
Heat of combustion.
Oxidation states.
Chemical stability. HOPE THIS HELPS!
