I think it is the third one.
Answer: 1.48 atmosphere
Explanation:
Pressure in kilopascal = 150
Pressure in atmosphere = ?
Recall that 1 atmosphere = 101.325 kilopascal
Hence, 1 atm = 101.325 kPa
Z atm = 150 kPa
To get the value of Z, cross multiply
150 kPa x 1 atm = 101.325 kPa x Z
150 kPa•atm = 101.325 kPa•Z
Divide both sides by 101.325 kPa
150 kPa•atm/101.325 kPa = 101.325 kPa•Z/101.325 kPa
1.48 atm = Z
Thus, 150 kPa is equivalent to 1.48 atmospheres
<u>Answer:</u>
Steep slopes on the contour map are identified using contour lines which are closely spaced.
<u>Explanation:</u>
Contour map also known as topographic maps are used to represent the three-dimensional portion of the earth’s surface in two-dimensional space. This map is used to represent the surface of the land such as steep, slopes, valleys. The contour maps are used in geological studies to understand the configuration of the earth's surface. Terms like map scale, vertical scale, contour lines are used on the contour map. Elevations are represented using contour lines. Contour lines placed very close to each represent the steep slopes and contour lines that are spaced farther away from each other represent the gentle slopes.
Answer:
31,380 Joules
Explanation:
Given Data:
Mass = m = 100 g
Temperature 1 = = 25 °C
Temperature 2 = = 100 °C
Specific Heat Constant = c = 4.184
Change in Temp. = ΔT = 100 - 25 = 75 °C
Required:
Heat = Q = ?
Formula:
Q = mcΔT
Solution:
Q = (100)(4.184)(75)
Q = 31, 380 Joules
Hope this helped!
~AH1807
Answer:
C. at low temperature and low pressure.
Explanation:
- <em>Le Châtelier's principle </em><em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
<em />
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.</em>
<em></em>
<em><u>Effect of pressure:</u></em>
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.
<em>So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), </em><em>so the equilibrium partial pressure of CO (g) can be maximized at low pressure.</em>
<em></em>
<u><em>Effect of temperature:</em></u>
- The reaction is exothermic because the sign of ΔH is (negative).
- So, we can write the reaction as:
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.</em>
- Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, <em>so the equilibrium partial pressure of CO (g) can be maximized at low temperature.</em>
<em></em>
<em>C. at low temperature and low pressure.</em>
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