Decomposition reaction D. 2H20⇒ 2H2 + O2
<h3>Further explanation
</h3>
Every chemical reaction involves a compound consisting of reactants and products
Reactants are compounds that react and form new compounds called products
There are several forms of reactions that can occur, including single replacement, double replacement, synthesis, decomposition, etc.
A. 2C2H6 + 702 ⇒ 4CO2 + 6H20
Combustion : reaction of Hydrocarbon and Oxygen
B. AgNO3 + LiCl → AgCl + LINO3
Double replacement : there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product
C. Ca + MgS → CaS + Mg
Single replacement :one element replaces the other elements of a compound to produce new elements and compounds
D. 2H20⇒ 2H2 + O2
Decomposition : One compound breaks down into 2 components
Answer:
a. 2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
b. 0.957 g
Explanation:
Step 1: Write the balanced equation
2 HgO(s) ⇒ 2 Hg(l) + O₂(g)
Step 2: Convert 130.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15
K = 130.0°C + 273.15
K = 403.2 K
Step 3: Calculate the moles of O₂
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 1 atm × 0.0730 L/0.0821 atm.L/mol.K × 403.2 K
n = 2.21 × 10⁻³ mol
Step 4: Calculate the moles of HgO that produced 2.21 × 10⁻³ moles of O₂
The molar ratio of HgO to O₂ is 2:1. The moles of HgO required are 2/1 × 2.21 × 10⁻³ mol = 4.42 × 10⁻³ mol.
Step 5: Calculate the mass corresponding to 4.42 × 10⁻³ moles of HgO
The molar mass of HgO is 216.59 g/mol.
4.42 × 10⁻³ mol × 216.59 g/mol = 0.957 g
<u>Answer:</u>
<u>For a:</u> The edge length of the unit cell is 314 pm
<u>For b:</u> The radius of the molybdenum atom is 135.9 pm
<u>Explanation:</u>
To calculate the edge length for given density of metal, we use the equation:
where,
= density = 
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal (molybdenum) = 95.94 g/mol
= Avogadro's number = 
a = edge length of unit cell =?
Putting values in above equation, we get:
![10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm](https://tex.z-dn.net/?f=10.28%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%20%28a%29%5E3%7D%5C%5C%5C%5Ca%5E3%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%2010.28%7D%3D3.099%5Ctimes%2010%5E%7B-23%7D%5C%5C%5C%5Ca%3D%5Csqrt%5B3%5D%7B3.099%5Ctimes%2010%5E%7B-23%7D%7D%3D3.14%5Ctimes%2010%5E%7B-8%7Dcm%3D314pm)
Conversion factor used: 
Hence, the edge length of the unit cell is 314 pm
To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:
where,
R = radius of the lattice = ?
a = edge length = 314 pm
Putting values in above equation, we get:
Hence, the radius of the molybdenum atom is 135.9 pm
Answer:
There are lots of methods.
Explanation:
Usually, animals like pandas live a shorter lifespan in the wild than in captivity. A little fact, there is only one brown panda in the entire world, so it would be very, very rare to see one. The Smithsonian National Zoo, for example, are working to protect pandas, as well as other species.
Democritus, a Greek philosopher, first developed the idea of atoms (around 460 B.C., I believe).
Hope this helps!
