Answer:
Change in temperature of calorimeter is 
Explanation:
Molar mass of anethole = 148.2 g/mol
So, 0.950 g of anethole = 
of anethole = 0.00641 moles of anethole
1 mol of anethole releases 5541 kJ of heat upon combustion
So, 0.00641 moles of anethole release 
of heat or 35.52 kJ of heat
7.854 kJ of heat increases 
temperature of calorimeter.
So, 35.52 kJ of heat increases 
or temperature of calorimeter
So, change in temperature of calorimeter is
Answer:
The hardness is more than 2.5. Hope this helps.
B, D and E
Explanation:
conversion factors
1c = 8oz
1pt = 2c
1qt = 2pt
For A and B
ounces to cup = 160/8 = 20c
cup to pints = 20c / 2c = 10pt
pint to quarts = 10pt/2pt = 5qt
B applies as four 1-quart and two 1-pt = 5 1-quart
Considering C
4 8oz = 32oz
160-32 = 128oz /8 = 16c/2 = 8pints C does not apply
considering D
8 * 8 = 64oz
160 - 64 = 96oz/8 = 12c/2 = 6pt/2 = 3qt
So D applies
E applies
Answer:
Option B. Plutonium-243
Explanation:
To know the the correct answer to the question, we shall write the balanced equation for the reaction showing curium-247 (²⁴⁷Cm) emitting alpha particle. This is illustrated below:
²⁴⁷₉₆Cm —> ᵐₙX + ⁴₂He
Next, we shall determine m, n and X. This can be obtained as follow:
247 = m + 4
Collect like terms
247 – 4 = m
243 = m
Thus, m is 243
96 = n + 2
Collect like terms
96 – 2 = n
94 = n
Thus, n is 94
m = 243
n = 94
ᵐₙX => ²⁴³₉₄X => ²⁴³₉₄Pu
Thus, the equation is:
²⁴⁷₉₆Cm —> ²⁴³₉₄Pu + ⁴₂He
From the illustrations above, the daughter nuclei formed is Plutonium-243 (²⁴³₉₄Pu)
When water vapor in the atmosphere loses heat and cools down, condensation happens. As the water vapor cools down and condenses, it attaches to small particles of dust floating in the atmosphere, forming tiny liquid water droplets.
