The formula units in the substances are as follows:
- Br2 = 8.99 × 10^23 formula units
- MgCl2 = 1.51 × 10^24 formula units
- H2O = 2.57 × 10^24 formula units
- Fe = 2.57 × 10^24 formula units
<h3>How many moles are in 239.2 g of the given substances?</h3>
The moles of the substances are determined from their molar mass.
Molar mass of the substances is given as follows:
- Br2 = 160 g/mol
- MgCl2 = 95 g/mol
- H2O = 18 g/mol
- Fe = 56 g/mol
Formula units = mass/molar mass × 6.02 × 10^23
The formula units in the substances are as follows:
- Br2 = 239.2/160 × 6.02 × 10^23 = 8.99 × 10^23 formula units
- MgCl2 = 239.2/95 × 6.02 × 10^23 = 1.51 × 10^24 formula units
- H2O = 239.2/18 × 6.02 × 10^23 = 2.57 × 10^24 formula units
- Fe = 239.2/56 × 6.02 × 10^23 = 2.57 × 10^24 formula units
In conclusion, the number of formula units is derived from the moles and Avogadro number.
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So your answer would pretty much be 2.80 x 10^24. The picture is just the explanation and how you would get that answer.
In general chemistry, isotopes are substances that belong to one specific element. So, they all have the same atomic numbers. But they only differ in the mass numbers, or the number of protons and neutrons in the nucleus. In a nutshell, they only differ in the number of neutrons.
For Nickel, there are 5 naturally occurring isotopes. Their identities, masses and relative abundance are listed below
Isotope Abundance Atomic Mass
Ni-58 68.0769% <span>57.9353 amu
Ni-60 </span>26.2231% <span>59.9308 amu
Ni-61 </span>1.1399 % <span>60.9311 amu
Ni-62 </span>3.6345% <span>61.9283 amu
Ni-64 </span>0.9256% <span>63.9280 amu
To determine the average atomic mass of Nickel, the equation would be:
Average atomic mass = </span>∑Abundance×Atomic Mass
Using the equation, the answer would be:
Average atomic mass = 57.9353(68.0769%) + 59.9308(26.2231%) + 60.9311(1.1399%) + 61.9283(3.6345%) + 63.9280(0.9256%)
Average atomic mass = 58.6933 amu
The image of the bonds are missing, so i have attached it.
Answer:
A) - Sigma bond
-Sp³ and Sp³
- None
B) - Sigma and pi bond
- Sp² of C and p of O
- p of C and P of O
Explanation:
A) For compound 1;
- the molecular orbital type is sigma bond due to the end-to-end overlapping.
- Atomic orbitals in the sigma bond will be Sp³ and Sp³
- Atomic orbitals in the pi bond would be nil because there is no pi bond.
B) For compound 2;
- the molecular orbital type is sigma and pi bond
-Atomic orbitals in the sigma bond would be Sp² of C and p of O
- The Atomic orbitals in the pi bond will be; p of C and p of O
A temperature change in a reaction indicates a chemical change