Answer:
306.6g/mol
Explanation:
To calculate molecular weight you need to know how many grams ou have in a determined amount of moles of substance. As you have the mass of the sample (1.42g), you need to find how many moles are, as follows:
The reaction of the acid HX with the base YOH is:
HX + YOH → H₂O + YX
<em>1 mole of acid reacts per mole of base.</em>
<em />
In a titration, the solution turned pink when moles base = moles acid.
Moles of base that the student added (Using the volume and molarity of the solution) are:
32.48mL = 0.03248L ₓ (0.1426 moles base / L) = 0.004632 moles of base
As the titration is in equivalence point, there are 0.004632 moles of the acid
Molecular weight (Ratio between grams of sample and its moles) is:
1.42g / 0.004632 moles =
<h3>306.6g/mol</h3>
Answer:
A- Small ice pellets that may fall to the ground in a mixture of rain and snow in the form of a solid
Explanation:
Small ice pellets that may fall to the ground in a mixture of rain and snow in the form of a solid best describes hail.
Answer:
Chemical equation:
HNO₃ + Al(OH)₃ → Al(NO₃)₃ + H₂O
Explanation:
Chemical equation:
HNO₃ + Al(OH)₃ → Al(NO₃)₃ + H₂O
Balanced chemical equation:
3HNO₃ + Al(OH)₃ → Al(NO₃)₃ + 3H₂O
Ionic equation:
3H⁺ + 3NO⁻₃(aq) + Al(OH)₃(s) → Al³⁺(aq) + 3NO₃⁻¹(aq) + 3H₂O(l)
Net ionic equation:
Al(OH)₃(s) + 3H⁺(aq) → Al³⁺(aq) + 3H₂O(l)
The NO⁻₃ are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.
Spectator ions:
These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.
Explanation:
it helps in when the kidney has used the needful nutrients been taken then the unwanted waste product it sends it to the nervous system then the nervous system passes the waste product to the bladder in form of urine
Answer:
20 drops of ethanol dispensed by dropper will be 1.0 mL ethanol.
Explanation:
Mass of 15 drops of ethanol = 0.60 g
Mass of 1 drop of an ethanol = 
Mass of 1.0 mL of ethanol = m
Volume of ethanol = 1.0 mL
Density of an ethanol = d = 0.80 g/ml
Number of drops in 0.80 g of ethanol = x
20 drops of ethanol dispensed by dropper will be 1.0 mL ethanol.
