Answer:5 moles ofCarbonmonoxide and 3.5 moles of oxygen gas.This in combine to yield carbon dioxide.
Explanation:
5C + 6O2----------5CO + 7/2O2.
When carbon combine with oxygen, carbon monoxide is formed first and it later recombine with oxygen to yield carbon dioxide.
Answer:
mass of HCl = 243.5426 grams
Explanation:
1- we will get the mass of the reacting gold:
volume of gold = length * width * height
volume of gold = 3.2 * 3.8 * 2.8 = 34.048 cm^3 = 34.048 ml<span>
density = mass / volume
Therefore:
mass = density * volume
mass of gold = </span>19.3 * 34.048 = 657.1264 grams
2- we will get the number of moles of the reacting gold:
number of moles = mass / molar mass
number of moles = 657.1264 / 196.96657
number of moles = 3.3362 moles
3- we will get the number of moles of the HCl:
First, we will balanced the given equation. The balanced equation will be as follows:
Au + 2HCl ......> AuCl2 + H2
This means that one mole of Au reacts with 2 moles of HCl.
Therefore 3.3362 moles will react with 2*3.3362 = 6.6724 moles of HCL
4- we will get the mass of the HCl:
From the periodic table:
molar mass of H = 1 gram
molar mass of Cl = 35.5 grams
Therefore:
molar mass of HCl = 1 + 35.5 = 36.5 grams/mole
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass of HCl = 6.6724 * 36.5
mass of HCl = 243.5426 grams
Hope this helps :)
Answer:
65.2L
Explanation:
Using the general gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (Litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (Kelvin)
According to the information provided in this question,
P = 1.631 atm
V = ?
n = 4.3 moles
T = 28°C = 28 + 273 = 301K
Using PV = nRT
V = nRT/P
V = 4.3 × 0.0821 × 301 ÷ 1.631
V = 106.26 ÷ 1.631
V = 65.15
Volume of the gas = 65.2L
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
Answer:
The answer to your question is: CH₄ + 3/2 O₂ ⇒ CO₂ + 2 H₂O
Explanation:
Methane = CH₄
Oxygen = O
Carbon dioxide = CO₂
Water = H₂O
CH₄ + 3/2 O₂ ⇒ CO₂ + 2 H₂O
This is the balanced equation