Answer:
The common thing is the compound water
Explanation:
in condensation h2O is expelled while in hydrolysis water is used or added
B. Hydra, because Hydra<span>are an ideal class of </span>Cnidaria<span> to research and on which to run tests.</span>
Answer:
0.032 L or 32 mL
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 9.0 M
V1 = This is what we're looking for.
M2 = 0.145 M
V2 = 2 L
Solve for V1 --> V1 = M2V2/M1
V1 = (0.145 M)(2 L) / (9.0 M) = 0.032 L
Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
First, the symbol for sodium oxide is Na₂O
Each Na (sodium) has a charge of 1+, and each O has a charge of 2- :
Na₂¹⁺O²⁻
There are two Na's, however, and each one is 1+, however, so the Na₂ has a total charge of 2+. Because of this, the 2+ from the 2 Na's and the 2- from the O cancel each other out to make 0.
