Answer:
compound, but I could be wrong
Whereas semidiurnal tides are observed at the equator at all times, most locations north or south of the equator experiencetwo unequal high tides and twounequal low tides per tidal day; this is called a mixed tide and the difference in height between successive high (or low) tides iscalled the diurnal inequality.
setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :

Q<K⇒The reaction moved to the right (products)
Setup 2 :

Q=K⇒the system at equilibrium
Setup 3 :

Q>K⇒The reaction moved to the left (reactants)
Answer:
mass
Explanation:
It is the amount of matter .
test tube
Explanation:
you state it is small and a test tube is the smallest container out of the answers you listed