<span>Balanced chemial equation:
2NaI(aq)+Hg2(NO3)2(aq) →Hg2 I2 (s) + 2 NaNO3 (aq)
You can see it better if I use latex:
As per the phases this is the interpretation:
The symbols (aq) stands for aquous meaning that the compound is dissolved in water.
The symbol (s) stands for solid, meaning tha the compound precipitate and is not dissolved in water.</span><span>
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Answer:
0.57 M
Explanation:
rate = change in concentration /time
Initial concentration of Cl2O5 = 1.16 M
Let the concentration of Cl2O5 after 5.70 seconds be y
rate = (1.16 - y)/5.7
The reaction follows a first order
Therefore, rate = ky = 0.184y
0.184y = (1.16 - y)/5.7
0.184y × 5.7 = 1.16 - y
1.0488y + y = 1.16
2.0488y = 1.16
y = 1.16/2.0488 = 0.57 M
Concentration of Cl2O5 after 5.70 seconds is 0.57 M
Drill cores from the ocean floor were dated and found to be very young compared to the age of the earth. This means the crust had to be formed recently, which can be explained by creation of crust at a spreading center.
Answer:
126.0g of water were initially present
Explanation:
The electrolysis of water occurs as follows:
2H₂O(l) ⇄ 2H₂(g) + O₂(g)
<em>Where 2 moles of water produce 2 moles of hydrogen and 1 mole of oxygen.</em>
<em />
To find the mass of water we need to determine moles of oxygen and hydrogen, thus:
<em>Moles Hydrogen:</em>
14.0g H₂ ₓ (1mol / 2g H₂) = 7 moles H₂
<em>Moles Oxygen:</em>
112.0g O₂ ₓ (1mol / 32g) = 3.5 moles O₂
Based on the chemical equation, the moles of water initially present were 7 moles (That produce 7 moles H₂ and 3.5 moles O₂). The mass of 7 moles of H₂O is:
7 moles H₂O * (18g / mol) =
<h3>126.0g of water were initially present</h3>
Answer:
222.30 L
Explanation:
We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:
Mass of NH₃ = 100 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 100 / 17
Mole of NH₃ = 5.88 moles
Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:
N₂ + 3H₂ —> 2NH₃
From the balanced equation above,
3 moles of H₂ reacted to produce 2 moles NH₃.
Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e
Xmol of H₂ = (3 × 5.88)/2
Xmol of H₂ = 8.82 moles
Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:
Pressure (P) = 95 KPa
Temperature (T) = 15 °C = 15 + 273 = 288 K
Number of mole of H₂ (n) = 8.82 moles
Gas constant (R) = 8.314 KPa.L/Kmol
Volume (V) =?
PV = nRT
95 × V = 8.82 × 8.314 × 288
95 × V = 21118.89024
Divide both side by 95
V = 21118.89024 / 95
V = 222.30 L
Thus the volume of Hydrogen needed for the reaction is 222.30 L
