Oxygen is the element. It has a total of 6 valence electrons.
quá trình oxi hóa: Mg0 -> Mg2+ + 2e-
quá trình khử: 2H+ + 2e- -> H2 (cần 2 H+ để tạo 1 H2)
----> chỉ cần thêm 2 vào trước HCl (vì H+ là do HCl cung cấp)
----> Mg + 2HCl -> MgCl2 + H2
yay :)
<span>54.8 g of MgI2 can be produced.
To solve this, you need to determine the molar mass of each reactant and the product. First, look up the atomic weights of iodine and magnesium
Atomic weight of Iodine = 126.90447
Atomic weight of Magnesium = 24.305
Molar mass of MgI2 = 24.305 + 2 * 126.90447 = 278.11394
Now determine how many moles of Iodine and Magnesium you have
moles of Iodine = 50.0 g / 126.90447 g/mol = 0.393997154 mole
moles of Magnesium = 5.15 / 24.305 g/mol = 0.211890557 mole
Since for every magnesium atom, you need 2 iodine atoms and since the number of moles of available iodine isn't at least 2 times the available moles of magnesium, iodine is the limiting reagent.
So figure out how many moles of magnesium will be consumed by the iodine
0.393997154 mole / 2 = 0.196998577 mole.
This means that you can make 0.196998577 moles of MgI2. Now simply multiply by the previously calculated molar mass of MgI2
0.196998577 mole * 278.11394 g/mole = 54.78805 g
Round the result to the correct number of significant figures.
54.78805 g = 54.8 g</span>
Answer:
λ = 5.68×10⁻⁷ m
Explanation:
Given data:
Energy of photon = 3.50 ×10⁻¹⁹ J
Wavelength of photon = ?
Solution:
E = hc/λ
h = planck's constant = 6.63×10⁻³⁴ Js
c = 3×10⁸ m/s
Now we will put the values in formula.
3.50 ×10⁻¹⁹ J = 6.63×10⁻³⁴ Js × 3×10⁸ m/s/ λ
λ = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 3.50 ×10⁻¹⁹ J
λ = 19.89×10⁻²⁶ J.m / 3.50 ×10⁻¹⁹ J
λ = 5.68×10⁻⁷ m
Answer:
Lost pigment of marker when dipped in alcohol
Explanation:
dependent viable = output
so it's the output of what happens after the input.
- she put the marker in the water which is the independent variable, that's the input
- the output or the result of that decision is having lost pigment in the marker
