132 g of C , 22 g of H , 176 g of O
132 + 22 + 176 => 330 g <span>of the substance
</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u H = 1.0 u O = 16.0 u
C = 132 / 12.0 => 11 moles
H = 22 / 1.0 => 22 moles
O = 176 / 16.0 => 11 moles
Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1
H = 22 / 11 => 2
O = 11 / 11 => 1
formula empirically <span>is : CH</span>₂O
hope this helps!
Bases- soap, baking soda
Acids- oranges,lemons
idk
This reaction would give rise to two products.
- 2-bromo-3-methylhexane, and
- 3-bromo-3-methylhexane.
However, 2-bromo-3-methylhexane would be more common than 3-bromo-3-methylhexane among the products.
The hydrogen atom in a hydrogen bromide molecule carries a partial positive charge. It is attracted to the double bond region with a high electron density. The hydrogen-bromine bond breaks when HBr gets too close to a double bond to produces a proton
and a bromide ion
.
The proton would attack the double bond to produce a carbocation. It could attach itself to either the second or the third carbon atom.
Carbocations are unstable and might decompose over time. The first carbocation is more stable than the second for having three alkyl groups- i.e., straight carbon chains- attached to the center of the positive charge. Alkyl groups have stabilizing positive induction effect on positively-charged carbon. The second carbocation has only two, and is therefore not as stable. The first carbocation thus has the greatest chance to react with a bromide ion to produce a stable halocarbon.
Bromide ions are negatively charged. They attach themselves to carbocations at the center of positive charge. Adding a bromide ion to the first carbocation would produce 3-bromo-3-methylhexane whereas adding to the second produces 2-bromo-3-methylhexane.
The <em>most likely</em> product of this reaction is therefore 3-bromo-3-methylhexane.
Answer:
Thermocline separate these two layers.
Explanation:
Ocean consist of three major layers on the basis of temperature.
1. Upper layer
2. Deep layer
3. Thermocline
Upper layer:
The first layer is called upper layer which is present on the surface and directly expose to the sun. The sun heat up this upper layer easily and warm it.
Deep layer
The deep layer is present below the thermocline. It is present in deep where sunlight can not approach to it and its temperature remain low.
Thermocline
It is the middle layer, present between the upper and deep layer. It separated the upper layer from deep layer. Its upper portion is warmer while lower is colder.
The volume of nitrogen monoxide that occupy at STP is= 277 Ml
calculation
The volume is obtained using the combined gas law that is P1V1/T1= P2V2?
Where P1 = 720 MmHg
V1 = 400ml
T1= 100 +273= 373 K
At STP temperature = 273 K and pressure= 760 mm hg
therefore T2= 273 k
p2 = 760 mmhg
V2=?
make V2 the subject of the formula
V2= (T2 ×P1 ×V1)/(P2×T1)
V2 is therefore = (273k x720 mmhg x 400 ml)/(760 mmhg x373K) = 277 Ml