Answer:
5.06atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
V1 = initial volume (Litres)
V2 = final volume (Litres)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 1.34 atm
P2 = ?
V1 = 5.48 L
V2 = 1.32 L
T1 = 61 °C = 61 + 273 = 334K
T2 = 31 °C = 31 + 273 = 304K
Using P1V1/T1 = P2V2/T2
1.34 × 5.48/334 = P2 × 1.32/304
7.34/334 = 1.32P2/304
Cross multiply
334 × 1.32P2 = 304 × 7.34
440.88P2 = 2231.36
P2 = 2231.36/440.88
P2 = 5.06
The final pressure is 5.06atm
The answer is Cl because its oxidation number goes from a +7 down to a -1.
Answer:
5.45*10⁻⁴ moles of silane gas (SiH₄) are present in 8.68 mL measured at 18°C and 1.50 atm.
Explanation:
An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P * V = n * R * T
In this case:
- P= 1.5 atm
- V= 8.68 mL= 0.00868 L (being 1000 mL= 1 L)
- n= ?
- R= 0.082

- T= 18 C= 291 K (being 0 C= 273 K)
Replacing:
1.5 atm* 0.00868 L= n* 0.082
*291 K
Solving:

n= 5.45*10⁻⁴ moles
<u><em>5.45*10⁻⁴ moles of silane gas (SiH₄) are present in 8.68 mL measured at 18°C and 1.50 atm.</em></u>
I need more information... what was the experiment??
Answer: The mass of air present in the room is 37.068 kg
Explanation : Given,
Length of the room = 10.0 ft
Breadth of the room = 11.0 ft
Height of the room = 10.0 ft
To calculate the volume of the room by using the formula of volume of cuboid, we use the equation:

where,
V = volume of the room
l = length of the room
b = breadth of the room
h = height of of the room
Putting values in above equation, we get:

Conversion used : 
Now we have to calculate the mass of air in the room.



Conversion used : (1 kg = 1000 g)
Therefore, the mass of air present in the room is 37.068 kg