For multiple covalent bonds to form in molecules, the molecules must contain carbon nitrogen or oxygen.
<u>Explanation:</u>
- Think about carbon dioxide (CO2). If every oxygen atom imparts one electron to the carbon molecule, there will be 6 electrons in carbon particle and 7 electrons in every oxygen atom. This doesn't give the carbon atom as a total octet.
- Sometimes more than one set of electrons is shared between two atoms. In carbon dioxide, a second electron from every oxygen atom is likewise imparted with the central carbon atom, and the carbon particle imparts one more electron with every oxygen atom.
- Two sets of electrons shared between two atoms make a double bond between the atoms. A few particles contain triple bonds, covalent bonds in which three sets of electrons are shared by two atoms.
C
He needs an battery to make the simple electromagnet
Answer is 56 protons and 56 electrons.
<em>Explanation;</em>
Atomic number is equal to number of protons. Hence, when the atomic number is 56, it means that atom has 56 protons.
When the element is in neutral state, number of protons = number of electrons. Hence, we can say that barium atom has 56 electrons.
But same element can have different number of neutrons. Those are called isotopes. Hence, we cannot say that there are 56 neutrons in barium atom without having its mass number. (Mass number = number of protons + number of neutrons)
Answer:
A tendon is a fibrous connective tissue that attaches muscle to bone. Tendons may also attach muscles to structures such as the eyeball. A tendon serves to move the bone or structure.
Explanation:
Answer:
91383 J
Explanation:
The equation of the reaction can be represented as:
------>
Given that:
The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.
The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.
= 
where:
= enthalpy of reaction
= the difference in the heat capacities of the products and the reactants.
∴
= 
= ![1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'](https://tex.z-dn.net/?f=1%2891300%20J.mol%5E%7B-1%7D%20%29%20%2B%5Cint%5Climits%5E%7B435%7D_%7B298.15%7D%20%5B%7B%2829.86%29-%5Cfrac%7B1%7D%7B2%7D%2829.38%29-%5Cfrac%7B1%7D%7B2%7D29.13%7D%5DJ.K%5E%7B-1%7D.mol%5E%7B-1%7D%20%5C%2C%20dT%27)
= 91300 J + (0.605 J.K⁻¹)(435-298.15)K
= 91382.79 J
≅ 91383 J
