Answer:
22Ω
Explanation:
Given parameters:
Potential difference = 3.3V
Current = 0.15A
Unknown:
Resistance = ?
Solution:
According to ohm's law, potential difference, current and resistance are related by the expression below;
V = I R
where V is the voltage
I is the current
R is the resistance
3.3 = 0.15 x R
R = 
= 22Ω
Answer:
Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. Therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.
Explanation:
That would be correct as stated.
The base units used in the metric system are meter for the measurement of distance or displacement, liter as unit measurement of volume, kilogram as a unit of measurement of mass and seconds as unit of measurement of time. The answer is here is D. gram.
Answer:
-177.9 kJ.
Explanation:
Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.
