Answer:
C-Its product is heavier than each of its reactants.
Explanation:
Correct equation:
¹⁴₇N + ¹₁H → ¹⁵₈O
In the reaction above, we can conclude within the given conditions of the reaction that the product formed is heavier than the reactants.
The product is oxygen with a mass number of 15 as shown by the superscript preceeding the symbol of the atom.
The reactants are:
Nitrogen, N with a mass number of 14
Hydrogen, H with a mass number of 1
The mass number is a true reflection of the mass of an atom. It clearly shows the mass of the nucleons which are the most massive particles that makes up an atom. The nucleons are protons and neutrons that makes up the tiny nucleus of the atom.
Oxygen here has more nucleons that each of Nitrogen and Hydrogen.
Answer:
Non-zero digits are always significant.
Any zeros between two significant digits are significant.
A final zero or trailing zeros in the decimal portion ONLY are significant. If a number ends in zeros to the right of the decimal point, those zeros are significant.
Explanation:
1.138 has 4 significant figures, which are 1, 1, 3 and 8. The numbers after the decimal point are decimals and are significant figures.
Answer:
Air & Water
Explanation:
Air and water is the common mixtures in the book of the book called "Science & Land"
Answer:
The correct answer is 0, 235 mol
Explanation:
We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol:
1 atm x 5, 25l = n x 0, 082 l atm / K mol x 273 K
n= 1 atm x 5, 25l /0, 082 l atm / K mol x 273 K
n= 0, 235 mol
You must verify that the number of atoms of each type is equal on both sides of the chemical equation: same number of C, same number of H and same number of O on both sides.
<span>A. C4H6 + 5.5O2 ---> 4CO2 + 3H2O
element reactant side product side
C 4 4
H 6 3*2 = 6
O 5.5 * 2 = 11 4*2 + 3 = 11
Then, this equation is balanced.
</span>Do the same with the other equations if you want to verify that they are not balanced.
Answer: option A.
