Answer:
1. (put in simpler words so no one knows you got it off of brainly)
The electricity that flows to our homes is generated in power stations. From here, it flows through large transmission lines, which carry it to substations. Finally, distribution lines carry electricity from substations to houses, businesses, and schools.
2. (put in simpler words so no one knows you got it off of brainly)
The transmission over long distances creates power losses. The major part of the energy losses comes from Joule effect in transformers and power lines. The energy is lost as heat in the conductors. This can be minimized by
answer choice 1: reducing technical losses including: replacing incorrectly sized transformers, improving the connection quality of conductors (power lines), and increasing the availability of reactive power by installing capacitor banks along transmission lines.
answer choice 2: (just in case answer choice 1 isn't what you're looking for)
You can reduce losses in your home by spreading out your electricity use evenly throughout the day, instead of running all your appliances at once.
Explanation:
hope this helped <3
A atomic number hope i helped
Answer:
My personal favorite is dragon ball super.
Explanation:
Answer:
4th line
Explanation:
-water cycle helps regulate the temperature on the Earth.
Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
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Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
