Answer:- 27.7 grams of 
are produced.
Solution:- The balanced equation is:
let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.
Molar mass of ![Pb(NO_3)_2 = 207.2+2(14.01)+6(16) = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2](https://tex.z-dn.net/?f=Pb%28NO_3%29_2%20%3D%20207.2%2B2%2814.01%29%2B6%2816%29%20%20%3D%20331.22%20gram%20per%20mol%3C%2Fp%3E%3Cp%3Emolar%20mass%20of%20NaI%20%3D%2022.99%2B126.90%20%3D%20149.89%20gram%20per%20mol%3C%2Fp%3E%3Cp%3EMolar%20mass%20of%20%5Btex%5DPbI_2)
= 207.2+2(126.90) = 461 gram per mol
let's do the calculations for the grams of the product for the given grams of each of the reactant:
= 
= 
From above calculations, NaI gives least amount of , so the answer is, 27.7 g of are produced.
Answer: In order to be alive there must be cells present and since sugar does not contain cells it is not alive.
Explanation:
In earths surface or the bottom of the Ocean
Answer:
Rb+
Explanation:
Since they are telling us that the equivalence point was reached after 17.0 mL of 2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.
Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n, of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.
Thus our calculations are:
V = 17.0 mL x 1 L / 1000 mL = 0.017 L
2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol
0.0425 mol = 4.36 g/ MW XOH
MW of XOH = (atomic weight of X + 16 + 1)
so solving the above equation we get:
0.0425 = 4.36 / (X + 17 )
0.7225 +0.0425X = 4.36
0.0425X = 4.36 -0.7225 = 3.6375
X = 3.6375/0.0425 = 85.59
The unknown alkali is Rb which has an atomic weight of 85.47 g/mol
Answer:
12.6.
Explanation:
- We should calculate the no. of millimoles of KOH and HCl:
no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.
no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.
- It is clear that the no. of millimoles of KOH is higher than that of HCl:
So,
[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.
∵ pOH = -log[OH⁻]
∴ pOH = -log(0.395 M) = 1.4.
∵ pH + pOH = 14.
∴ pH = 14 - pOH = 14 - 1.4 = 12.6.
