Answer:
58.61 grams
Explanation:
Taking The molecular weight of NaCl = 58.44 grams/mole
<u>Determine how many grams of NaCl to prepare the bath solution </u>
first we will calculate the moles of NaCl that is contained in 6L of 170 mM of NaCI solution
= ( 6 * 170 ) / 1000
= 1020 / 1000 = 1.020 moles
next
determine how many grams of NaCl
= moles of NaCl * molar mass of NaCl
= 1.020 * 58.44
= 58.61 grams
Answer:
Four moles of the cation
Explanation:
2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)
Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.
The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.
This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.
Answer:
429.4 kJ are absorbed in the endothermic reaction.
Explanation:
The balanced chemical equation tells us that 1168 kJ of heat are absorbed in the reaction when 4 mol of NH₃ (g) react with 5 mol O₂ (g).
So what we need is to calculates how many moles represent 25 g NH₃(g) and calculate the heat absorbed. (NH₃ is the limiting reagent)
Molar Mass NH₃ = 17.03 g/mol
mol NH₃ = 25.00 g/ 17.03 g/mol = 1.47 mol
1168 kJ /4 mol NH₃ x 1.47 mol NH₃ = 429.4 kJ
Here, we have to get the number of atoms present in the 100 plane of the FCC crystal lattice.
There will be 2 atoms in 100 plane of FCC crystal lattice.
In the face centered crystal (FCC) lattice there are atoms at each corner of the cube and each are shared by 4 another atoms. And an atom is present at the face of the crystal.
For the 100 plane of the Miller indices the intercepts are a, ∞, ∞ or 2a, ∞, ∞.
Thus, for the 4 atoms of the corner at the cube shared by 4 other atoms will contribute, 4 ×
= 1 and the un-shared atoms at the face will contribute another 1, which make the total atom 1 + 1 = 2.
The moles of oxygen that are produced when 26.5 mol of Al2O3 decomposes is 39.8 mol
<u>calculation</u>
<u> </u> 2Al2O3 + 4Al +3 O2
- use the mole ratio of Al2O3 to O2 to determine the moles of O2.
- that is from the equation above the mole ratio of Al2O3 : O2 is 2:3
- the moles of O2 is therefore=n 26.5 mol x3/2= 39.8 moles