Answer:
B. Q > K precipitate will form
Explanation:
The reaction is;
Ba(NO3)2(aq) + Na2CO3(aq) ------> BaCO3(s) + 2NaNO3(aq)
Hence the reaction could form a precipitate of BaCO3.
Number of moles of carbonate ions = 50/1000 * 0.10 M = 5 * 10^-3 moles
Number of moles of Barium ions = 20/1000 * 0.10 M = 2 * 10^-3 moles
Total volume after reaction = 20ml + 50ml = 70 ml or 0.07 L
Molarity Barium ions = 5 * 10^-3 moles/ 0.07 L = 0.07 M
Molarity carbonate ions = 2 * 10^-3 moles/ 0.07 L =0.03 M
Q = [Ba^2+] [CO3^2-] = 0.07 * 0.03 = 2.1 * 10^-3
But K = 2.58 × 10
^−
9
We can clearly see that Q>K therefore precipitate will form
Answer:
C.
Explanation:
The electronic configuration of N (7 electrons): 1s² 2s² 2p³.
The orbital 1s is filled with two electrons and their spinning direction is opposite and also electrons of 2s.
3p contains (3 electrons) should fill the 3 orbitals firstly. Every orbital contains 1 electron and be in the same spin direction.
So, the right choice is c.
A is wrong because 2 electrons of 3p are paired in the first orbital before filling every orbital.
B is wrong because the 2 electrons of 1s and 2s are in the same direction and also 2 electrons of 3p are paired in the first orbital before filling every orbital.
D is also wrong the 2 electrons of 1s and 2s are in the same direction and the electron in the second orbital of 3p are in opposite direction of the other 2 electrons.
Answer:
0.049 mol/L.s
Explanation:
The decomposition of hydrogen peroxide is:
![Rate = -\dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{\Delta [H_2O_2]}{\Delta t}= \dfrac{ 2 \Delta [H_2O_2]}{\Delta t}](https://tex.z-dn.net/?f=Rate%20%3D%20-%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D%3D%20%5Cdfrac%7B%202%20%20%5CDelta%20%5BH_2O_2%5D%7D%7B%5CDelta%20t%7D)
The rate of decomposition reaction = the rate of formation of 
= 0.098 mol/L.s
∴
Rate of formation of 
= 0.049 mol/L.s
Answer:
23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr
Explanation:
The balanced equation here is
6NaBr + 1AlO3 = 3Na2O + 2AlBr3
6 moles of NaBr are required to produce 2 moles of AlBr3
Mass of one mole of NaBr = 102.894 g/mol
Mass of one mole of AlBr3 = 266.69 g/mol
Mass of 6 moles of NaBr = 6*102.894 g/mol
Mass of two moles of AlBr3 = 2*266.69 g/mol
6*102.894 g NaBr produces 2*266.69 g of AlBr3
23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr
The statement that describes the chemical reaction is D chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromide<span>. The symbol "Cl" represents chlorine. The symbols in the brackets show the physical state of the substance, (g) is gaseous, (s) is solid, (aq) is aqueous and (l) is liquid.</span>
