Answer:
32.5g of sodium carbonate
Explanation:
Reaction of sodium carbonate (Na₂CO₃) with Mg²⁺ and Ca²⁺ as follows:
Na₂CO₃(aq) + Ca²⁺(aq) → CaCO₃(s)
Na₂CO₃(aq) + Mg²⁺(aq) → MgCO₃(s)
<em>1 mole of carbonate reacts per mole of the cations.</em>
<em />
To know the mass of sodium carbonate we must know the moles of carbonate we need to add based on the moles of the cations:
<em>Moles Mg²⁺:</em>
2.91L * (0.0661 moles MgCl₂ / 1L) = 0.192 moles MgCl₂ = Moles Mg²⁺
<em>Moles Ca²⁺:</em>
2.91L * (0.0396mol Ca(NO₃)₂ / 1L) = 0.115 moles Ca(NO₃)₂ = Moles Ca²⁺
That means moles of sodium carbonate you must add are:
0.192 moles + 0.115 moles = 0.307 moles sodium carbonate.
In grams (Using molar mass Na₂CO₃ = 105.99g/mol):
0.307 moles Na₂CO₃ * (105.99g / mol) =
<h3>32.5g of sodium carbonate</h3>
Answer:
2.2 x 10²² molecules.
Explanation:
- Firstly, we need to calculate the no. of moles in (6.0 g) sodium phosphate:
<em>no. of moles = mass/molar mass </em>= (6.0 g)/(163.94 g/mol) = <em>0.0366 mol.</em>
- <em>It is known that every mole of a molecule contains Avogadro's number (6.022 x 10²³) of molecules.</em>
<em />
<u><em>using cross multiplication:</em></u>
1.0 mole of sodium phosphate contains → 6.022 x 10²³ molecules.
0.0366 mole of sodium phosphate contains → ??? molecules.
<em>∴ The no. of molecules in 6.0 g of sodium phosphate</em> = (6.022 x 10²³ molecules)(0.0366 mole)/(1.0 mole) = <em>2.2 x 10²² molecules.</em>
Answer:
When we look at an arbitrary point in the sky, away from the sun, we see only the light that was redirected by the atmosphere into our line of sight. Because that occurs much more often for blue light than for red, the sky appears blue. Violet light is actually scattered even a bit more strongly than blue.
Explanation:
Answer:
A) = 4.7 × 10⁻⁴atm
Explanation:
Given that,
Kp = 1.5*10³ at 400°C
partial pressure pN2 = 0.10 atm
partial pressure pH2 = 0.15 atm
To determine:
Partial pressure pNH3 at equilibrium
The decomposition reaction is:-
2NH3(g) ↔N2(g) + 3H2(g)
Kp = [pH2]³[pN2]/[pNH3]²
pNH3 =√ [(pH2)³(pN2)/Kp]
pNH3 = √(0.15)³(0.10)/1.5*10³ = 4.74*10⁻⁴ atm
![K_p = \frac{[pH_2] ^3[pN_2]}{[pNH_3]^2} \\pNH_3 = \sqrt{\frac{(pH_2)^3(pN_2)}{pNH_3} } \\pNH_3 = \sqrt{\frac{(0.15)^3(0.10)}{1.5 \times 10^3} } \\=4.74 \times 10^-^4atm](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cfrac%7B%5BpH_2%5D%20%5E3%5BpN_2%5D%7D%7B%5BpNH_3%5D%5E2%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%28pH_2%29%5E3%28pN_2%29%7D%7BpNH_3%7D%20%7D%20%5C%5CpNH_3%20%3D%20%5Csqrt%7B%5Cfrac%7B%280.15%29%5E3%280.10%29%7D%7B1.5%20%5Ctimes%2010%5E3%7D%20%7D%20%5C%5C%3D4.74%20%5Ctimes%2010%5E-%5E4atm)
= 4.7 × 10⁻⁴atm
Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V
