H2SO4 + 2 NaOH ----> Na2SO4 + 2 H2O.
0.085 L * 0.176 mol/L = 0.01496 mol H2SO4
is neutralised by 0.01496 mol * 2
= 0.02992 mol NaOH.
1000 mL of 0.492 M NaOH
contains 0.492 moles NaPH.
0.02992 / 0.452 * 1000 mL
= 66.19 = 66 mL
Limiting reactant : O₂
Mass of N₂O₄ produced = 95.83 g
<h3>Further explanation</h3>
Given
50g nitrous oxide
50g oxygen
Reaction
2N20 + 302 - 2N204
Required
Limiting reactant
mass of N204 produced
Solution
mol N₂O :
mol O₂ :
2N₂O+3O₂⇒ 2N₂O₄
ICE method
1.136 1.5625
1.0416 1.5625 1.0416
0.0944 0 1.0416
Limiting reactant : Oxygen-O₂
Mass N₂O₄(MW=92 g/mol) :
Answer:
2-methyl-butene
Explanation:
For the E2 mechanism, we have an <u>anti-elimination</u>. The Br leaves the molecule and the base removes the hydrogen in the anti position to form the double that's why only one structure is produced. (See figure 1)
Since we have 2 hydrogens on the right carbon, we cannot indicate a <u>specific stereoisomer</u>, in other words, it is not possible to assign a <u>Z or E</u> configuration for this alkene.
The answer to this question would be: <span>thermal metamorphism
</span>
Metamorphism is a change in the mineral texture without causing the rock to become liquid/magma. In this case, the metamorphic change to the rock is caused by the heat energy or thermal energy of the magma. This kind of mechanism is also called contact mechanism as the thermal energy is transferred by contact so this question option is a bit ambiguous.
Hello!
To find the amount of atoms that are in 80.45 grams of magnesium, we will need to know Avogadro's number and the mass of one mole of magnesium.
Avogadro's number is 6.02 x 10^23 atoms, and one mole of magnesium is equal to 24.31 grams.
1. Divide by one mole of magnesium
80.45 / 24.31 = 3.309 moles (rounded to the number of sigfigs)
2. Multiply moles by Avogadro's number
3.309 x (6.02 x 10^23) = 1.99 x 10^24 (rounded to the number of sigfigs)
Therefore, there are 1.99 x 10^24 atoms in 80.45 grams of magnesium.
