Answer:
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Explanation:
If the net force acting on an object is 0 N, the forces are
Since the forces equal to 0, it's balanced because the object isn't moving.
Toluene is subjected to the action of the following reagents in the order given: (1) KMnO4,OH-, heat; then H3O (2) HNO3, H2SO4 (3) Br2, FeBr3 What is the final product of this sequence?
Answer:
See image attached
Explanation:
The reaction of toluene with alkaline potassium permanganate in the presence of heat leads to the oxidation of the -CH3 to give benzoic acid.
Reaction benzoic acid with HNO3/H2SO4 yields the nitronium ion (NO2+).
Recall that -COOH is a metal director and deactivated the ring towards electrophilic substitution hence the m-nitrobenzoic acid is formed.
Reaction with FeBr3/Br2 yields the product shown in the image attached.
Which statement describes how phase changes can be diagrammed as a substance is heated?
The phase is on the y-axis and the temperature is on the x-axis.
The temperature is on the y-axis and the phase is on the x-axis.
The time is on the y-axis and the temperature is on the x-axis.
The temperature is on the y-axis and the time is on the x-axis.
Answer:
The temperature is on the y-axis and the time is on the x-axis.
Explanation:
Which statement describes how phase changes can be diagrammed as a substance is heated? (D is the answer)
The phase is on the y-axis and the temperature is on the x-axis.
The temperature is on the y-axis and the phase is on the x-axis.
The time is on the y-axis and the temperature is on the x-axis.
The temperature is on the y-axis and the time is on the x-axis.
Which statement describes the appearance of a temperature-vs.-time graph? (C is the answer)
A horizontal line shows that the temperature increases at a constant rate over time.
A vertical line shows that the temperature decreases at a constant rate over time.
Horizontal lines where the temperature is constant during phase changes connect upward-sloping lines where the temperature increases.
Horizontal lines where the temperature increases are connected by upward-sloping lines where the temperature is constant for each phase.
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A weather balloon with a volume of 3.40774 L
is released from Earth’s surface at sea level.
What volume will the balloon occupy at an
altitude of 20.0 km, where the air pressure is
10 kPa?
Answer in units of L.
Answer: The volume occupied at an altitude of 20.0 km is 34.5289 L
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
where,
[tex]P_1[/tex] = initial pressure of gas = 101.325 kPa ( sea level)
[tex]P_2[/tex] = final pressure of gas = 10 kPa
[tex]V_1[/tex] = initial volume of gas = 3.40774 L
[tex]V_2[/tex] = final volume of gas = ?
Now put all the given values in the above equation, we get the final pressure of gas.
[tex]101.325\times 3.40774=10\times V_2[/tex]
[tex]V_2=34.5289L[/tex]
Therefore, the volume occupied at an altitude of 20.0 km is 34.5289 L
Which of these lead (II) salts will dissolve to the greatest extent in water?
a. PbSO4, Ksp = 1.7x10^-8
b. PbI2, Ksp = 6.5x10^-9
c. PbCrO4, Ksp = 1.8x10^-14
d. PbS, Ksp = 2.5x10^-27
e. Pb3(AsO4)2, Ksp = 4.0x10^-36
Answer:
a. PbSO4, Ksp = 1.7x10^-8.
Explanation:
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In this case, since the solubility product indicates how likely a solid is able to ionize and consequently dissolve in water, we can infer that the larger the solubility product Ksp, the more ions are able dissolve in water; therefore the proper answer goes with the largest Ksp, which is a. PbSO4, Ksp = 1.7x10^-8 since the power goes closer to 1 than the other options.
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How does the density and distribution of your “stars” change as the balloon expands?
Answer:
The Universe is constantly expanding and as it expands the stars and objects in space move farther apart, just like the points on the balloon when air is blown into it. Density and distribution of "stars" as the balloon expands because when volume increases the density will increase.
Hope this Helps
A gas at -20c occupies volume 140 ml calculate temperature at which the volume of the gas becomes 65 ml pressure constant
Answer:
–156 °C
Explanation:
The following data were obtained from the question:
Initial temperature (T1) = –20 °C
Initial volume (V1) = 140 mL
Final volume (V2) = 65 mL
Final temperature (T2) =?
Pressure = constant.
Next, we shall convert –20 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T1) = –20 °C
Initial temperature (T1) = –20 °C + 273 = 253 K.
Next, we shall determine the new temperature of the gas as follow:
Initial temperature (T1) = 253 K
Initial volume (V1) = 140 mL
Final volume (V2) = 65 mL
Final temperature (T2) =?
V1/T1 = V2/T2
140/253 = 65/T2
Cross multiply
140 × T2 = 253 × 65
140 × T2 = 16445
Divide both side by 140
T2 = 16445 /140
T2 = 117 K
Finally, we shall convert 117 K to celcius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T2 = 117 K
T2 = 117 K – 273
T2 = –156 °C
Thus, the new temperature of the gas is –156 °C