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Nana76 [90]
2 years ago
6

What is the name of Pb(NO3)2? Explain how you determined the bond type and the steps you used to determine the naming convention

for the compound. Please be specific :)
Chemistry
1 answer:
Elena L [17]2 years ago
5 0

Answer: Lead(II) nitrate but idk the rest

Explanation:

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What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2
LekaFEV [45]
Molar mass CaCl₂ = 111.0 g/mol

number of moles:

n = mass of solute / molar mass

n = 85.3 / 111.0

n = 0.7684 moles of CaCl₂

M = n / V

0.788 M = <span>0.7684 / V
</span>
V = 0.7684 / 0.788

V = 0.97512 L

hope this helps!


3 0
2 years ago
What is the element with the nlx notation of 5d2?
lesantik [10]
The element with the nlx notation of 5d2 in the periodic table is hafnium with a symbol of hf. It is a chemical element and has an atomic number 72. H<span>afnium chemically resembles zirconium and is found in many zirconium minerals but it is a complete contrast of zirconium which is also chemically similar to hafnium.</span>
7 0
2 years ago
The statement that percent yield can never be greater than theoretical yield is another example of the ________.
Gnom [1K]
We can rephrase the statement with a little more specificity in order to understand the answer here.

The mass of the products can never be more than the The mass that is expected.
3 0
2 years ago
Oxides of virtually every element are known. bromine, for example, forms several oxides when treated with ozone. suppose you all
FromTheMoon [43]
Answer is: empirical formula of the product is Br₂O₅.
Chemical reaction: x/2Br₂ + y/3O₃ → BrₓOy.
m(Br₂) = 1,250 g.
m(BrₓOy) = 1,876 g.
n(Br₂) = m(Br₂) ÷ M(Br₂).
n(Br₂) = 1,25 g ÷ 159,81 g/mol.
n(Br₂) = 0,0078 mol.
n(Br) = 2 · 0,0078 mol = 0,0156 mol.
m(O₃) = 1,876 g - 1,25 g = 0,626 g.
n(O₃) = 0,626 g ÷ 48 g/mol = 0,013 mol.
n(O) = 0,039 · 3 = 0,039 mol
n(Br) : n(O) = 0,0156 mol : 0,039 mol.
n(Br) : n(O) = 1 : 2,5.
5 0
2 years ago
Read 2 more answers
A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0
2 years ago
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