Answer:
Yes, because the light was the manipulated variable
Explanation:
<u>Answer:</u> The 
for the reaction is -1835 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:
The intermediate balanced chemical reaction are:
(1) 
( × 4)
(2) 
The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B4%5Ctimes%20%28-%5CDelta%20H_1%29%5D%2B%5B1%5Ctimes%20%5CDelta%20H_2%5D)
Putting values in above equation, we get:
Hence, the for the reaction is -1835 kJ.
Answer:
Double replacement
Precipitation reaction
Explanation:
You have the reaction:
REACTANTS PRODUCTS
BaCl₂ (aq) + Na₂SO₄ (aq) ⇒ 2NaCl (aq) + BaSO₄(s)
The general form of a double replacement reaction is the following:
AB + CD ⇒ CD + AB
The reactants basically, exchanged partners. In the case of your problem, Barium(Ba) and Sodium(Na) switched places. So this makes it a double-replacement reaction.
Now how do I know it is a precipitation reaction. A precipitation reaction occurs when two solutions combine and salt is formed. Salt is solid, so how do I know that's what occured? Look at your equation again:
BaCl₂ (aq) + Na₂SO₄ (aq) ⇒ 2NaCl (aq) + BaSO₄(s)
aq means aqueous (liquid)
s means solid
If you look at the product formed in the reaction, from two solutions, it formed a solid. So this is your clue as to why it is a precipitation reaction.
The appropriate answer is b. A metallic bond allows metals to conduct electricity. Metallic bonds are formed by atoms of metals in which the outer electrons of the atoms from a common electron cloud. In metallic bonding the atoms are not used up in the actual bond but are shared so they are free to move about to conduct electricity. Electrons in the other bond types are held in place in the molecule an are not free to move about so they cannot conduct electricity.
