If there is an increase in industrial activity, that means that more heat will be dissipated to the atmosphere in the form of carbon dioxide. Industrialization requires fuel to keep the processes on the go. At the end of the pipeline, the combustion of fuel would result to carbon dioxide released to the atmosphere. That's how it is contributing to the global climate change through the greenhouse effect.
Answer:
4.81 moles
Explanation:
The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.
Pressure at which gauge reads zero = 14.7 psi
Pressure read by the gauge = 988 psi
Total pressure = 14.7 + 988 psi = 1002.7 psi
Also, P (psi) = P (atm) / 14.696
Pressure = 1002.7 / 14.696 = 68.2297 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 4.81 moles
Answer:
FALSE
Since 0.385 < 0.526, the value for week 3 is accepted.
Explanation:
Qexp = (|Xq - Xₙ₋₁|)/w
where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data
First, the data are arranged in decreasing order, from highest to lowest:
3. 5.6
2. 5.1
8. 5.1
1. 4.9
6. 4.9
5. 4.7
7. 4.5
4. 4.3
Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3
Qexp = (|5.6 - 5.1|)/1.3 = 0.385
From tables, at 95% confidence level, for n = 8, Qcrit = 0.526
Since 0.385 < 0.526, the value for week 3 is accepted.
Proton gives nucleus positive charge P+
