Answer:
E = 165.75×10⁻²⁶ J
Explanation:
Given data:
Wavelength = 0.12 m
Energy of wave = ?
Solution:
Formula:
E = h c/λ
c = 3×10⁸ m/s
h = 6.63×10⁻³⁴ Js
Now we will put the values in formula.
E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 0.12 m
E = 19.89×10⁻²⁶ J.m / 0.12 m
E = 165.75×10⁻²⁶ J
Answer:
The ballance half reactions are:
Mg²⁺ + 2e⁻ → Mg
6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O
Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)
Explanation:
Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)
Let's see the oxidations number.
As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.
Mg in reactants, acts with +2, so the oxidation number has decreased.
This is the reduction, so it has gained electrons.
Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.
Let's take a look to half reactions:
Mg²⁺ + 2e⁻ → Mg
Si → SiO₃²⁻ + 4e⁻
In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:
6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O
If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.
2Mg²⁺ + 4e⁻ → 2Mg
Now, that they are ballanced we can sum the half reactions:
2Mg²⁺ + 4e⁻ → 2Mg
6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O
2Mg²⁺ + 4e⁻ + 6OH⁻ + Si → 2Mg + SiO₃²⁻ + 4e⁻ + 3 H₂O
Your answer is 160 grams I hope this helps
Answer:
124 g (3 sig figs)
or
124.011 g (6 sig figs
Explanation:
Step 1: Calculate g/mol for AgNO₃
Ag - 107.868 g/mol
N - 14.01 g/mol
O - 16.00 g/mol
107.868 + 14.01 + 16.00(3) = 169.878 g/mol
Step 2: Multiply 0.73 moles by molar mass
0.73 mol (169.979 g/mol)
124 grams of AgNO₃
Answer:
CaCl2 + H2O Ca2+(aq) + 2Cl-(aq)
Explanation:
CaCl2 + H2O Ca2+(aq) + 2Cl-(aq)
When CaCl2 is dissolved in H2O (water) it will dissociate (dissolve) into Ca+2 and Cl- ions.
The dissolution of calcium chloride is an exothermic process.
Explanation:
Butanol (С4Н9OH)
Structural formula is :-
CH3-CH2-CH2-CH2-OH
Answer:
[K2SO4] = 4,75x10⁻⁷M ; [K⁺] = 9.50x10⁻⁷M ; [SO4⁻²] = 4,75x10⁻⁷M
SO4⁻²: 0.045ppm ; K⁺: 0.037ppm
[SO4⁻²] = 4,70x10⁻⁷ F
Explanation:
Determine the equation
K2SO4 → 2K⁺ + SO4⁻²
Each mole of potassium sulfate generates two moles of potassium cation and one mole of sulfate anion
Molar mass K2SO4: 174.26 g/m
Moles of K2SO4: grams / molar mass
2.07x10⁻⁴g / 174.26 g/m = 1.18x10⁻⁶ moles
Molarity: Moles of solute in 1 L of solution
1.18x10⁻⁶ moles / 2.5 L = 4,75x10⁻⁷M (K2SO4)
K⁺ : 4,75x10⁻⁷M . 2 = 9.50x10⁻⁷M
SO4⁻²: 4,75x10⁻⁷ M
1 mol of K2SO4 has 2 moles of K and 1 mol of SO4
1.18x10⁻⁶ moles of K2SO4 has 1.18x10⁻⁶ moles of SO4 and 2.37x10⁻⁶ moles of K.
1.18x10⁻⁶ moles of SO4⁻² are 1.13x10⁻⁴ grams (moles. molar mass)
2.37x10⁻⁶ moles of K are 9.26x10⁻⁵ grams (moles. molar mass)
These grams are in 2.5 L of water, so we need μg/mL to get ppm
2.5 L = 2500 mL
1.13x10⁻⁴ grams SO4⁻² are 113.35 μg (1 μg = 1x10⁶ g)
9.26x10⁻⁵ grams K⁺ are 92.6 μg (1 μg = 1x10⁶ g)
113.35 μg /2500 mL = 0.045ppm
92.6 μg /2500 mL = 0.037ppm
Formal concentration of SO4⁻² :
Formality = Number of formula weight of solute / Volume of solution (L)
(1.13x10⁻⁴ grams / 96.06 g ) / 2.5 L = 4,70x10⁻⁷ F