250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze: Yes, because there is enough energy.
<h3>At what temperature would a sample of water freeze?</h3>
- Note from the Facilitator: At certain temperatures, water changes its condition due to temperature variations. At sea level, fresh water changes from a solid to a liquid at 32°F (0°C). Liquid water freezes at temperatures below 32°F (0°C); this temperature is known as the freezing point of water.
- The fact that a single water molecule cannot transform into a solid, liquid, or gas is the answer. These names refer to collective behaviors of water molecules rather than to individual molecules.
- For instance, the solid (ice) has a collection of molecules that are bound together and arranged in a predictable manner. That cannot be accomplished by a single molecule alone
250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze: Yes, because there is enough energy.
To learn more about water freezing, refer to:
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Electrolysis is the process of applying an electrical current to a solution in order to separate the components of that liquid phase solution on the basis of their charge. If we consider molten calcium hydroxide, the ions present will be:
Ca⁺²
OH⁻
The cations, the ones with a positive charge, will be reduced at the cathode. Therefore, calcium metal will be produced at the cathode. The anions will be oxidized at the anode, which means hydrogen gas will be produced at the anode.
Answer:
8
Explanation:
Its official chemical symbol is O, and its atomic number is 8, which means that an oxygen atom has eight protons in its nucleus
Answer:
As there must be twice as many lithium ions as sulphide ions, the 'numbers'
must be different.
It's 4:8. i.e. 4 sulphides around each lithium; and 8 lithiums around each
sulphur.
It's an antifluorite structure.
Answer:
3.99 g
Explanation:
The following data were obtained from the question:
Half life (t½) = 2 years.
Original amount (N₀) = 128 g
Time (t) = 10 years
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 years.
Decay constant (K) =.?
K= 0.693/t½
K = 0.693/2
K = 0.3465 year¯¹.
Finally, we shall determine the amount remaining after 10 years i.e the amount remaining when they arrive on Earth. This can be obtained as follow:
Original amount (N₀) = 128 g
Time (t) = 10 years
Decay constant (K) = 0.3465 year¯¹.
Amount remaining (N) =..?
Log (N₀/N) = kt/2.3
Log (128/N) = (0.3465 × 10)/2.3
Log (128/N) = 1.5065
Take the antilog of 1.5065
128/N = Antilog (1.5065)
128/N = 32.1
Cross multiply
128 = 32.1 × N
Divide both side by 32.1
N = 128/32.1
N = 3.99 g
Therefore, the amount remaining is 3.99 g
