First convert celcius to Kelvin.
20 + 273 = 293K
31 + 273 = 304K
Now we can set up an equation based on the information we have.
V1 = 5
P1 = 365
T1 = 293
V2 = 5
P1 = x
T2 = 304
The equation be: 
Now just solve.
1825/293 = 5x/304
Cross multiply.
554800 = 1465x
Divide both sides by 1465
x = 378.7030717 which can then be rounded to 378.7 mmHg
Benzaldehyde or C6H5CHO would not undergo the aldol condensation because it does not contain an alpha-hydrogen in its structure. Aldol condensation is a type of reaction that happens between an enolate and an aldehyde or ketone leading to a alkene that has a planar structure. The lack of an alpha-hydrogen would not allow for it to undergo such process since it cannot enolize. Benzaldehyde undergoes a nucleophilic reaction known as Claisen-Schmidt condensation. It has somehow same mechanism of the aldol reaction however, the nucleophilic attack on the carbonyl happens even without the alpha-hydrogen but with an enolate that is from a ketone.
Answer:
2 HCl + CaCO3 = CaCl2 + H2O + CO2
Explanation:
On the right side we have 2xCl, to get the same amount we will multiply HCl on the left side by 2.
This will give us 2xH on the left side; we also have 2xH on the right side ( in H2O).
On both sides we have 1x Ca; 1x C.
On the left side we have 3XO (in CaCO3), on the right side we have 1O in H20 and 2xO in CO2, so 3x O on the right side as well.
The reaction is balanced.
Polar and non polar bonds only have to do with non metals. non polar have to do with non metal diatomic molecules while polar bonds deal with non metals that have different electronegativities
You are given
300 grams of FeCr2O7. You are required to get the mass of O2 in grams. The first
thing to do is to find a reaction between FeCr2O7 and O2. The balanced chemical
reaction is 2Fe + 6Cr + 21/2O2 -> Fe2(Cr2O7)3. The molar mass of Fe2(Cr2O7)3
is 759.65 grams per mole and 32 grams per mole for O2.
<span>
300g Fe2(Cr2O7)3
(1 mol Fe2(Cr2O7)3/759.65 g Fe2(Cr2O7)3)(21/2 mol O2/1 mole Fe2(Cr2O7)3)(32 g
O2/1 mole O2) = <u>132.7 g O2</u></span>
