Answer:
The percentage composition of the elements of the compound in the three samples is the same.
Explanation:
<em>The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements in the same proportion by mass.</em>
Sample A:
Mass of A = 4.31 g; mass of Z = 7.70 g
Total mass of sample = 12.01
Percentage mass of A in the sample = (4.31 * 100)/12.01 = 35.9 %
Percentage mass of Z in the sample = (7.70 * 100)/12.01 = 64.1 %
Sample B:
Percentage mass of A in the sample = 35.9 %
Percentage mass of Z in the sample = 64.1 %
Sample C:
Mass of A = 0.718 g; Total mass of sample = 2.00 g
mass of Z = mass of sample - mass of A = 2.00 g - 0.718 g = 1.282 g
Percentage mass of A in the sample = (0.718 * 100)/2.00 = 35.9 %
Percentage mass of Z in the sample = (1.282 * 100)/2.00 = 64.1 %
From the calculations, it can be seen that the percentage composition of the elements in the compound is the same for the three samples.
Answer:
Electron-pair geometry: tetrahedral
Molecular geometry: trigonal pyramidal
Hybridization: sp³
sp³ - 4 p
Explanation:
There is some info missing. I think this is the original question.
<em>For NBr₃, What are its electron-pair and molecular geometries? What is the hybridization of the nitrogen atom? What orbitals on N and Br overlap to form bonds between these elements?</em>
<em>The N-Br bonds are formed by the overlap of the ___ hybrid orbitals on nitrogen with ___ orbitals on Br.</em>
<em />
Nitrogen is a central atom surrounded by 4 electron domains. According to VESPR, the corresponding electron-pair geometry is tetrahedral.
Of these 4 electron domains, 3 represent covalent bonds with Br and 1 lone pair. According to VESPR, the corresponding molecular geometry is trigonal pyramidal.
In the nitrogen atom, 1 s orbital and 3 p orbitals hybridize to form 4 sp³ orbitals for each of the electron domains.
The N-Br bonds are formed by the overlap of the sp³ hybrid orbitals on nitrogen with 4p orbitals on Br.
Answer:
Explanation:
The gas ideal law is
PV= nRT (equation 1)
Where:
P = pressure
R = gas constant
T = temperature
n= moles of substance
V = volume
Working with equation 1 we can get
The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.
or
(equation 2)
The cylindrical container has a constant pressure p
The volume is the volume of a cylinder this is
Where:
r = radius
h = height
(pi) = number pi (3.1415)
This cylinder has a radius, r and height, h so the volume is
Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:
Replacing these values in the equation 2 we get:
(equation 2)
c- it says I need to write at least 20 words to submit it so- here haha
Answer is: 1,92 mol/L·s.
Chemical reaction: 2D(g) + 3E(g) + F(g) → <span>2G(g) + H(g).
</span>H is increasing at 0,64 mol/L·<span>s.
From chemical reaction n(H) : n(E) = 1 : 3.
0,64 mol : n(E) = 1 : 3.
n(E) = 1,92 mol.
</span>E is decreasing at 1,92 mol/L·s.
