In a voltaic cell made of a Ag strip and an Al strip immersed in 1.0 M aqueous AgNO_3 and Al(NO_3)_3 solutions, respectively, wi
1 answer:
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The structure of Alanine is shown below,
Except the carbon atom of carbonyl group which is Sp² Hybridized, all remaining atoms are Sp³ Hybridized. The hybridization of each element is depicted in a box below,
Except the carbon atom of carbonyl group which is Sp² Hybridized, all remaining atoms are Sp³ Hybridized. The hybridization of each element is depicted in a box below,
Answer:
1.822 g of magnesium hydroxide would be produced.
Explanation:
Balanced reaction:
Compound Molar mass (g/mol)
NaOH 39.997
95.211
58.3197
So, 2.50 g of NaOH = mol of NaOH = 0.0625 mol of NaOH
4.30 g of =
mol of
= 0.0452 mol of
According to balanced equation-
2 mol of NaOH produce 1 mol of
So, 0.0625 mol of NaOH produce mol of NaOH or 0.03125 mol of NaOH
1 mol of produces 1 mol of
So, 0.0452 mol of produce 0.0452 mol of
As least number of moles of are produced from NaOH therefore NaOH is the limiting reagent.
So, amount of would be produced = 0.03125 mol
= g
= 1.822 g
<u>Answer:</u> The mass difference between the two is 7.38 grams.
<u>Explanation:</u>
To calculate the number of moles, we use the equation given by ideal gas follows:
where,
P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)
V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)
T = Temperature =
R = Gas constant =
n = number of moles = ?
Putting values in above equation, we get:
To calculate the number of moles, we use the equation:
.....(1)
- <u>For air:</u>
Moles of air = 0.297 moles
Average molar mass of air = 28.8 g/mol
Putting values in equation 1, we get:
Mass of air, = 8.56 g
- <u>For helium gas:</u>
Moles of helium = 0.297 moles
Molar mass of helium = 4 g/mol
Putting values in equation 1, we get:
Mass of helium, = 1.18 g
Calculating the mass difference between the two:
Hence, the mass difference between the two is 7.38 grams.