<h2>
Question
</h2>
A sample of methane collected when the temp was 30 C and 760mmHg measures 398 mL. What would be the volume of the sample at -5 C and 616 mmHg pressure
<h2>
Answer:</h2>
434.32mL
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Explanation:</h2>
Using the combined gas law:
= k
Where;
P = Pressure
V = Volume
T = Temperature
k = constant.
It can be deduced that:
= 
= k ---------------------(i)
Where:
P₁ and P₂ are the initial and final pressures of the given gas
V₁ and V₂ are the initial and final volumes of the given gas
T₁ and T₂ are the initial and final temperatures of the gas.
<em>From the question:</em>
the gas is methane
P₁ = 760mmHg
P₂ = 616mmHg
V₁ = 398mL
V₂ = ?
T₁ = 30°C = (30 +273)K = 303K
T₂ = -5°C = (-5 +273)K = 268K
Substitute these values into equation (i) as follows;
= 
Solve for V₂
V₂ = 
V₂ = 434.32mL
Therefore, the volume of the sample at -5C and 616mmHg pressure is 434.32mL
Answer: The aluminium trihalide assists in the heterolytic bond fission of the bromine molecule.
Explanation:
The electrophilic aromatic bromination of acetalinide with molecular bromine requires the formation of Br+. This specie is formed when an Aluminum trihalide reacts with bromine. The Br-Br bond breaks heterolytically such that charged species are produced. This Br+ participates in the electrophilic aromatic substitution and AlX4- is formed where X4 shows a tetrahalide complex of aluminium
A melting point of over 700 C and a density of less than 2 g/cm3 can be observed for many group 2 elements. In this group, the density increases on moving down the group, whereas the melting point increases upto calcium and then starts decreasing.
Calcium, symbol Ca is the element with melting point around 840 C and density of 1.55 g/cm3 which is closest to the specified data range .
Answer:
There are three rules on determining how many significant figures are in a number:
Non-zero digits are always significant.
Any zeros between two significant digits are significant.
A final zero or trailing zeros in the decimal portion ONLY are significant.
Answer:
0.20 moles
Explanation:
The pressure is proportional to the quantity of gas at a given temperature and volume. So, the quantity needs to be increased by a factor of ...
(35 psi)/(29.2 psi) = 175/146 ≈ 1.19863
The fractional increase required is ...
1.19863 -1 = 0.19863
__
The quantity of air currently in the tire is ...
1 mol·519.67°R/(atm·23.6442 L) × (29.2/14.7 atm) × (11.6 L) / (45+459.67)°R
= 1.0035 mol
so we need to add ...
(fraction to add) × (current quantity) = amount to add
0.19863 × 1.0035 mol = 0.1993 mol = amount to add
About 0.20 moles of air must be added to the tire to bring the pressure up.
