Answer:
4Ba(CO3) -> 4BaO2 + 2CO2
Explanation:
I looked at the oxygens to balance this. Ba(CO3) normally has 3 oxygens. BaO2 and CO2 have 4 oxygens total. The common multiple of 3 & 4 is 12. So there should be 12 oxygens on both sides. Then I just found the coefficients that would give 12 oxygens on both sides and can balance the rest of the atoms.
<span>Among the given choices, the third option is the only one which illustrates single replacement.
(3)H2SO4 + Mg --> H2 + MgSO4
A single replacement is also termed as single-displacement reaction, a reaction by which an element in a compound, displaces another element.
It can be illustrated this way:
X + Y-Z → X-Z + Y</span>
C. both a and b
If a light bulb can last longer with the same amount of energy it is given, that means it can use less energy to do the same job compared to one that does not last longer with the same amount of energy it is given. It is much like how a more fuel efficient car will be able to go farther on the same tank of gas, but if you pair it with a car that doesn't have as great of an mpg, when they go the same distance, the car with the greater mpg spends less fuel.
If you don't have to use the energy when you aren't utilizing it, then you can conserve the energy for when you do need it.
Answer:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK
Explanation:
The temperature of an object can be written using different temperature scales.
The two most important scales are:
- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).
- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.
The expression to convert from Celsius degrees to Kelvin is:
Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK
