Answer:
im pretty sure the electron configuration of the barium ion Ba+2 is [Kr]5s24d105p6
Explanation:
Answer:
a. 5.36x10⁻⁴ g/mL
b. 4.29x10⁻⁵ g/mL
Explanation:
As the units for concentration are not specified, I'll respond using g/mL.
a. We <em>divide the sample mass by the final volume</em> in order to <u>calculate the concentration</u>:
- 0.268 g / 500 mL = 5.36x10⁻⁴ g/mL
b. We can use C₁V₁=C₂V₂ for this question:
- 8.00 mL * 5.36x10⁻⁴ g/mL = C₂ * 100.00 mL
Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
<span>1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced</span>
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2
<span>4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms</span>
Answer:
Here's what I've found
Explanation:
Refer to the attachment !
