Answer:
1400000 cm
Explanation:
14 km equals to 1400000 cm (14 km = 1400000 cm)
PLEASE MARK ME AS THE BRAINLIEST... ;)
Answer:
10.3 g Al
Explanation:
To find the excess mass of Al, you need to (1) convert grams I₂ to moles I₂ (via molar mass), then (2) convert moles I₂ to moles AlI₃ (via mole-to-mole ratio from equation coefficients), then (3) convert moles AlI₃ to grams AlI₃ (via molar mass). Now that you have the actual amount of AlI₃ produced, you need to (4) convert grams AlI₃ to moles AlI₃ (via molar mass), then (5) convert moles AlI₃ to moles Al (via mole-to-mole ratio from equation coefficients), and then (6) convert moles Al to grams Al (via molar mass). Now that you know the amount of Al actually needed to produce the product, you need to (7) find the excess mass of Al.
Molar Mass (Al): 26.982 g/mol
Molar Mass (I₂): 2(126.90 g/mol)
Molar Mass (I₂): 253.8 g/mol
Molar Mass (AlI₃): 26.982 g/mol + 3(126.90 g/mol)
Molar Mass (AlI₃): 407.682 g/mol
2 Al(s) + 3 I₂(g) -------> 2 AlI₃(g)
113 g I₂ 1 mole 2 moles AlI₃ 407.682 g
------------- x ---------------- x ---------------------- x ------------------- = 121 g AlI₃
253.8 g 3 moles I₂ 1 mole
121 g AlI₃ 1 mole 2 moles Al 26.982 g
--------------- x ------------------ x --------------------- x ----------------- = 8.01 g Al
407.682 g 2 moles AlI₃ 1 mole
Starting Amount - Mass Needed = Excess
18.3 g Al - 8.01 g Al = 10.3 g Al
<u>Answer:</u> The standard electrode potential of the cell is 0.62 V.
<u>Explanation:</u>
For the given chemical equation:
The substance having highest positive potential will always get reduced and will undergo reduction reaction.
<u>Oxidation half reaction:</u> ( × 3 )
<u>Reduction half reaction:</u> ( × 2 )
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the of the reaction, we use the equation:
Hence, the standard electrode potential of the cell is 0.62 V.
Molarity HNO3 = 0.85 M
Volume HNO3 = 38.5 mL in liters: 38.5 / 1000 => 0.0385 L
Molarity Ba(OH)2 = ?
Volume Ba(OH)2 = 20.0 mL / 1000 => 0.02 L
Number of moles HNO3 :
n = M x V
n = 0.85 x 0.0385<span> => 0.032725 moles
mole ratio :
</span><span>Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
</span>
1 mole Ba(OH)2 ----------- 2 moles HNO3
( moles Ba(OH)2 ) -------- 0.032725 HNO3
moles Ba(OH)2 = 0.032725 x 1 / 2
moles Ba(OH)2 = 0.032725 / 2
= 0.0163625 moles of Ba(OH)2
Molarity Ba(OH)2 :
M = n / V
M =0.0163<span>625</span> / <span>0.02
= 0.82 M
hope this helps!</span>
Answer:
Explanation:the formula for osmotic pressure is given in the first line, all the terms are explained before solution, from the question,
Molar mass=Mm=386.6
Mass= m= 14.6
Volume= 263/1000 = 0.263dm^3
Temperature=T= 298K.
Gas constant .R= 0.082
Then substitute into the equation πV= nRT
Making π subject of formula
π= nRT/V
Then further substitution