Answer: This is an oxidation-reduction (redox) reaction:
3 C-II - 12 e- → 3 CII (oxidation)
4 CrVI + 12 e- → 4 CrIII (reduction)
C2H5OH is a reducing agent, K2Cr2O7 is an oxidizing agent.
Answer:
False.
Explanation:
<u>The given statement asserts a false claim because the equator is the region that receives maximum sunlight</u>. The equator is placed right below the sun and thus, it tends to receive the maximum radiation across the year. While the poles are the coldest regions of the Earth because due to Earth's titled axis, they receive very few sun rays for a certain time of the year. Thus, <u>if we move away from the equator, we are likely to receive less radiation from the sun as the sun keeps getting farther while moving away from the equator</u>.
You can use grams to moles and moles to grams. In your case just grams to moles. So since you're given grams, you would divide that by the molar mass of CO2 because that's how many grams are in one mole. The mass for Carbon is 12.0104 g/mol and Oxygen it's 15.9994 g/mol so to find the molar mass you would add 12.0104 + (2*15.9994) which gives you a molar mass of 44.0095 g/mol. You divide your given mass (132g) by the molar mass, so there's 2.9993 moles or approximately 3 moles in 132 g of CO2.
Answer:
(i) ΔU = 116 J
(ii) ΔU = 289 J
(iii) ΔU = 1 KJ
(iv) ΔU = 0 J
(v) ΔU = 3.25 KJ
Explanation:
first law:
(i) W = 153 J; Q = - 37 J ( Q ( - ), losing friction )
⇒ ΔU = 153 - 37 = 116 J
(ii) W = 289 J; Q = 0 ( insulated)
⇒ ΔU = W = 289 J
(iii) Q = 1 KJ , W = 0 ( isovolumetric process)
⇒ ΔU = Q = 1 KJ
(iv) isothermal ( constant temperature )
∴ ΔT = 0° ( isothermal )
⇒ ΔU = 0 J
(v) isobaric ( constant pressure )
⇒ ΔU = Q + W
∴ Q = 15.6 KJ
∴ W = - ∫ P dV = - P ΔV; W (-) the system performs a job and the volume increases
.
∴ P = 950 KPa * ( 1000 Pa / KPa ) = 950000 Pa = 950000 J/m³
∴ ΔV = 18 - 5 = 13 L * ( m³ / 1000 L ) = 0.013 m³
⇒ W = - ( 950000 J/m³) * ( 0.013 m³ ) = - 12350 J ( - 12.35 KJ )
⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )
⇒ ΔU = 3.25 KJ
