Answer:
In an endothermic reaction, the products have more stored chemical energy than the reactants. In an exothermic reaction, the opposite is true. The products have less stored chemical energy than the reactants. The excess energy in the reactants is released to the surroundings
Answer:
The answers to your questions are given below.
Explanation:
__ NaBr + CaF₂ —> __ NaF + CaBr₂
The above equation can be balance as follow:
NaBr + CaF₂ —> NaF + CaBr₂
There are 2 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 2 before NaF as shown below:
NaBr + CaF₂ —> 2NaF + CaBr₂
There are 2 atoms of Na on the right side and 1 atom on the left side. It can be balance by writing 2 before NaBr as shown below:
2NaBr + CaF₂ —> 2NaF + CaBr₂
Now, the equation is balanced.
Atom >>>> Reactant >>>> Product
Na >>>>>> 2 >>>>>>>>>>> 2
Br >>>>>>> 2 >>>>>>>>>>> 2
Ca >>>>>> 1 >>>>>>>>>>>> 1
F >>>>>>> 2 >>>>>>>>>>>> 2
Answer:
A) H₂O at 120°C
Explanation:
It is possible to think the higher temperature, the greatest degree of disorder. That is because with a high temperature, vibrations of molecules increases.
In general, at low temperatures, the molecules are in solid state (The lowest degree of disorder), increasing its temperature, molecules becomes in liquids, and, with more temperature, are gases (The greatest degree of disorder).
Thus, the sample that has the greatest degree of disorder is:
<h3>A) H₂O at 120°C</h3>
You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.
For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.
6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2
For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.
13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2
As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.
In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!
<h2>Step 1 : Identify the given </h2>
Volume = 250mL
Density = 1.19 g/ML
<h2>Step 2 . Calculate the mass of HCL </h2>
Density = mass/volume
∴Mass = Density * Volume
= 1.19g/mL* 250mL
= 297,5g
<h2>Step 3 : Calculate the total mass of the solution, given that concentration HCL is 38% </h2>
Mass of the total solution can be calculated by the following :
38% = Mc /297.5 * 100
Mc = 38/100 *297.5
= 113.05grams
• Finally, this means that mass of the total solution of 0.125M HCL i,s 113grams, ,you would use this mass to prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0%
