<span>11.6 g of lead sulfide.
First, get the molar masses of lead and sulfur
Lead = 207.2
Sulfur = 32.065
Now determine how many moles of each we have avaiable
lead = 10.0 g / 207.2 g/mol = 0.048262548 mol
sulfur = 1.6 g / 32.065 g/mol = 0.049898643 = mol
This tells me that the what's being produced is PbS instead of PbS2 and that there's a very slight excess of sulfur in the original reaction. So on the 2nd reaction with the same amount of lead and twice the amount of sulfur, there will be an even greater excess of sulfur and that you'll get 11.6 g of lead sulfide.</span>
Close to about -36 degrees Celsius ( - 33 degrees Fahrenheit) in January but can drop to -60 degrees Celsius ( - 76 degrees Fahrenheit).
in July, it can be close to -19 degrees Celsius (- 2 degrees Fahrenheit)
Answer:
Rb
Explanation:
Ionization energy is defined required energy to eliminate or remove an electron from an ion or atom.
From the given elements Rb or Rubidium has the lowest ionization energy as it has lowest shielding effect, so it easy to remove electron from it's shell.
The ionization energy generally decreases from top to bottom in groups due to lower shielding effect and outer electrons are loosely packed so easy to remove and increases from left to right across a period because of valence shell stability.
Rubidium has atomic number 37 and lies below than other given elements and also placed in the left side in the periodic table.
Hence, the correct option is Rb
Answer:
C2H6 + 2 Cl2 = C2H4Cl2 + 2 HCl
Explanation:
First, we write the reaction equation:
2KI + PbNO₃ → K₂NO₃ + PbI₂
The molar ratio of KI to PbNO₃ is 2 : 1
Moles of PbNO₃ present:
Moles = concentration (M) x volume (dm³)
= 0.194 x 0.195
= 0.038
Moles of KI required = 2 x 0.038 = 0.076 moles
concentration = moles / volume
volume = moles / concentration
= 0.076 / 0.2
= 0.38 L = 380 ml
