Answer : The metal used was iron (the specific heat capacity is ).
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of unknown metal = ?
= specific heat of water =
= mass of unknown metal = 150 g
= mass of water = 200 g
= final temperature of water =
= initial temperature of unknown metal =
= initial temperature of water =
Now put all the given values in the above formula, we get
Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).
Therefore, the metal used was iron (the specific heat capacity is ).
A general exponential expression is something like:
A^n
This means that we need to multiply the number A by itself n times.
Using that we will get (-2)^6 = 64
With that definition, we can rewrite:
(-2)^6 = (-2)*(-2)*(-2)*(-2)*(-2)*(-2)
So we just need to solve the above expression.
Also, remember the rule of signs:
(-)*(-) = (+)
We will get:
(-2)*(-2)*(-2)*(-2)*(-2)*(-2) = [(-2)*(-2)]*[(-2)*(-2)]*[(-2)*(-2)]
= 4*4*4 = 16*4 = 64
Then we got:
(-2)^6 = 64
If you want to learn more, you can read:
brainly.com/question/17172630
Answer:
D. concentration, as the concentrations of reactants and products remain unchanged after equilibrium is reached.
Explanation:
I put it for the test and i got right hehe
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
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)
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)