The lateral line is a system of sense organs found in aquatic vertebrates, used to detect movement, vibration, and pressure gradients in the surrounding water. ... For example, fish can use their lateral line system to follow the vortices produced by fleeing prey
Answer:
is the isotopic notation of the atom
Explanation:
The isotope notation is:
![^a_bX](https://tex.z-dn.net/?f=%5Ea_bX)
<em>Where a is the mass number = Number of protons + Number of neutrons</em>
<em>b is atomic number = Number of protons</em>
The atomic number define the nature of the atom, the element with atomic number = 15 is phosphorus, P:
![^a_bP](https://tex.z-dn.net/?f=%5Ea_bP)
a = 15 protons + 16 neutrons = 31
b = 15
is the isotopic notation of the atom.
the correct answer is option ( A ) Mixture
Explanation:
It is defined as the substance that is made by the combination of two or more different components. Or, we can say that it is created when solids liquids or gases mixed with one another. Hence, the correct option is, mixture.
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Answer:
option D= Gold (I) nitride
Explanation:
The name of the given compound is gold(I) nitride.
Molar mass can be determine by following way:
molar mass Au3N = (molar mass of gold × 3) + (molar mass of nitrogen)
molar mass Au3N = (196.97 × 3 ) + ( 14 )
molar mass of Au3N = 590.91 g/mol + 14 g/mol
molar mass of Au3N = 604.91 g/mol
The nitrogen has valency of -3 so three Au(+1) will require while the valency of Au is (1+) one nitrogen will require to make the compound overall neutral.
Au3N
3(1+) + (-3) = 0
+3 - 3 = 0
0 = 0
The overall charge is 0, the compound will be neutral.
The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,
2H2 + O2 --> 2H2O
The masses of each of the reactants are calculated below.
2H2 = 4(1.01 g) = 4.04 g
O2 = 2(16 g) = 32 g
Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
(1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2
Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.
<em>Answer: 1.22 g of oxygen</em>