Answer:
( c ) sunlight.
Explanation:
the leaves are closer to the sunlight they require.
The atomic mass would be 28.08535 amu. Multiply 27.9769 by .92297 = 25.803. Multiply 28.9765 by .046832 to get 1.357. Multiply 29.9738 by .03872 to get .925351136. Add 25.803 + 1.357 + .03872 to get 28.08535 amu
Hey there!:
Molar mass:
H2 = 2.01 g/mol ; H2O = 18.01
Given the reaction:
2 H2 + O2 = 2 H2O
2 * (2.01 ) g H2 ------------- 2 * ( 18.01 ) g H2O
mass H2 --------------------- 1.80 g H2O
mass H2 = 1.80 * 2 * 2.01 / 2* 18.01
mass H2 = 7.236 / 36.02
mass H2 = 0.2008 g
Hope that helps!
Answer:
1.35 g
Explanation:
Data Given:
mass of Potassium Permagnate (KMnO₄) = 3.34 g
Mass of Oxygen: ?
Solution:
First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)
So,
Molar Mass of KMnO₄ = 39 + 55 + 4(16)
Molar Mass of KMnO₄ = 158 g/mol
Calculate the mole percent composition of Oxygen in Potassium Permagnate (KMnO₄).
Mass contributed by Oxygen (O) = 4 (16) = 64 g
Since the percentage of compound is 100
So,
Percent of Oxygen (O) = 64 / 158 x 100
Percent of Oxygen (O) = 40.5 %
It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.
So,
for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be
mass of Oxygen (O) = 0.405 x 3.34 g
mass of Oxygen (O) = 1.35 g
Answer:
0.1082M of Barium Hydroxide
Explanation:
KHP reacts with Ba(OH)2 as follows:
2KHP + Ba(OH)2 → 2H2O + Ba²⁺ + 2K⁺ + 2P²⁻
<em>Where 2 moles of KHP reacts per mole of barium hydroxide</em>
<em />
To solve this question we must find the moles of KHP in 1.37g. With these moles and the reaction we can find the moles of Ba(OH)2 and its molarity using the volume of the solution (31.0mL = 0.0310L) as follows:
<em>Moles KHP -Molar mass: 204.22g/mol-</em>
1.37g * (1mol / 204.22g) = 0.006708 moles KHP
<em>Moles Ba(OH)2:</em>
0.006708 moles KHP * (1mol Ba(OH)2 / 2mol KHP) =
0.003354 moles Ba(OH)2
<em>Molarity:</em>
0.003354 moles Ba(OH)2 / 0.0310L =
<h3>0.1082M of Barium Hydroxide</h3>
