Answer:
HCl
Explanation:
A limiting reactant is the lowest amount of reactant so that the reaction will stop after all that reactant used.
To answer this question, you have to change all reactant units into moles. You have 52g HCl and its molecular mass is 36.46g/mole. The number of HCl in moles will be: 52g/ (36.46g/mole)= 1.43 moles.
In the balanced reaction formula, you can see that you need one mole of HCl and one mole of NaOH for each reaction. Both molecule's coefficient is 1.
If we used up all NaOH, then the number of HCl left will be:
(moles of NaOH used * NaOH coefficient) - (moles of HCl used * HCl coefficient)
(moles of HCl you have * HCl coefficient) - (moles of NaOH used * NaOH coefficient)
(1.43 moles *1 ) - (2.5 moles*1 ) = -1.07 mole
Since the result is minus, then it means we need more HCl and we can't use all NaOH.
If we used up all the HCl, then the number of NaOH left will be:
(moles of NaOH you have* NaOH coefficient) - (moles of HCl used * HCl coefficient)
(2.5 moles*1 )- (1.43 moles *1 )= 1.07 moles
Since the result is plus, then it means we can use all HCl. Then HCl is the limiting reactant
P2O5 = Phosphorus pentoxide
CuO = Copper (II) oxide
NH4CI = Ammonium Chloride
Mn(OH)2 = Pyrochroite
H2O2 = Hydrogen peroxide
P4S9 = Tetraphosphorus nonasulfide
CIO2 = Chlorine dioxide
NaF = Sodium fluoride
FeSO3 = Iron (II) Sulfite
Fe(NO3)3 = Iron (III) Nitrate
Cr(NO2)3 = Chromium (III) Nitrite
NaHCO3 = Sodium Hydrogen Carbonate
H2PO4 = Dihydrogen Phosphate Ion
NaCN = Sodium Cyanide
IF7 = Iodine Heptafluoride
PCI3 = Phosphorus Trichloride
Answer:
answer
<u>Cold Front - a zone separating two air masses, of which the cooler, denser mass is advancing and replacing the warmer. </u>
<u>Warm Front - a transition zone between a mass of warm air and the cold air it is replacing. </u>
<u>Stationary Front - a front between warm and cold air masses that is moving very slowly or not at all.</u>
Answer: The thermocline begins at 100 meters of depth.
Step-to-step explanation:-
Thermocline is a transition oceanic water layer between deep and surface water in which water temperature decreases rapidly with increasing depth.
From the given graph it cam be seen that at 100 meters the thermoline begins such that the temperature drops from 
to 
.
Hence, the thermocline begins at 100 meters of depth.
Answer:
The answer to your question is V2 = 4.97 l
Explanation:
Data
Volume 1 = V1 = 4.40 L Volume 2 =
Temperature 1 = T1 = 19°C Temperature 2 = T2 = 37°C
Pressure 1 = P1 = 783 mmHg Pressure 2 = 735 mmHg
Process
1.- Convert temperature to °K
T1 = 19 + 273 = 292°K
T2 = 37 + 273 = 310°K
2.- Use the combined gas law to solve this problem
P1V1/T1 = P2V2/T2
-Solve for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (783 x 4.40 x 310) / (292 x 735)
-Simplification
V2 = 1068012 / 214620
-Result
V2 = 4.97 l
