The residential end-use sector has the largest seasonal variance, with significant spikes in demand every summer and winter. Virtually all homes that have air conditioning use electricity as the main source of cooling in the summer, while winter heating needs are met by a variety of fuels. Some homes use electric resistance heating and electric heat pumps, but even homes with other heating fuels such as natural gas or fuel oil still use some electricity to power furnace fans, boiler circulation pumps, and compressors.
The commercial sector experiences less variance in electricity use, although it shows a noticeable increase in the summer and a slight increase in the winter. Compared to the residential sector, a smaller portion of commercial sector energy consumption is devoted to heating, cooling, and ventilation. However, other energy fuels beyond electricity can be used in the commercial sector to meet both heating and cooling needs. For example, some commercial buildings use natural gas-fired chillers for cooling.
The industrial sector's demand for electricity is relatively flat (with just a slight increase in the summer) because a much smaller portion of its energy consumption (electric and otherwise) is used for heating and cooling. Economic variables generally play a larger role in industrial energy use than weather-related factors. However, seasonal changes can affect industrial activity. For example, in the refining industry, different seasonal slates of petroleum products as well as different seasonal processes may affect electricity needs.
When a system experiences a disturbance ( such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state.
The different types of microscopes are all necessary because not all experiments require the same level of magnification. For dissections low magnification is sufficient, so a dissecting microscope works very well, while for viewing single cells the 1000 fold magnification of a compound light microscope is more accurate.
Answer:
The correct option is OA.
C2H4O2 + NaHCO3 - NaC2H302 + H2O + CO2
Explanation:
To solve this you have to check the number of elements in both sides of the equation.
Answer:
2.47L
Explanation:
Using the combined gas law equation as follows:
P1V1/T1= P2V2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 705mmHg
P2 = 760mmHg (STP)
V1 = 3.00L
V2 = ?
T1 = 35°C = 35 + 273 = 308K
T2 = 273K (STP)
Using P1V1/T1= P2V2/T2
705 × 3/308 = 760 × V2/273
2115/308 = 760V2/273
Cross multiply
308 × 760V2 = 2115 × 273
234,080V2 = 577,395
V2 = 577,395 ÷ 234,080
V2 = 2.47L