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2025 WAEC/NECO(GCE), JUPEB HQπŸŽ“πŸŽ“

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AGRIC PRATICAL ANSWERS BELOW 1(a): 1. Soil fertility: Earthworms require a favourable soil environment that provides sufficient food and moisture to support their livelihood. 2. Soil structure: Earthworms prefer soil that is porous, allowing for adequate air and water circulation, while also providing a proper balance of nutrients. 1(b): 1. Soil aeration: Earthworms burrow through the soil, creating channels that allow air and water to circulate through the soil more easily. This improves the soil's aeration and drainage. 2. Nutrient cycling: Earthworms help to break down organic matter, such as fallen leaves and dead roots, into nutrients that can be used by plants. This process is known as nutrient cycling. 3. Soil structure: Earthworms can improve soil structure by creating burrows that help to loosen the soil, allowing for better root penetration and greater uptake of nutrients. 4. Water retention: Earthworms help to create pore spaces in the soil which can hold onto water. This can help to prevent soil erosion and improve plant growth. 5. Biological activity: Earthworms are an important part of the soil food web. They consume decaying organic matter and provide food for other soil organisms, such as bacteria and fungi, which help to improve soil health and fertility. ======================================= ======================================= Chat Us Now With N500 Airtel Recharge Card Or Bank Transfer To Subscribe And Get Full Answers Always Subscribe To Get Questions And Answers On Time Before Exam Start. Keep Inviting your Friends And Classmates To Eagle Eye Exam Solution 100% Legit. Tell the whole world about us Eagle Eye Exam Solution Team We remain your favorite source. Ensure you Subscribe for your next paper. ===================================== JOIN OUR OFFICIAL TELEGRAM CHANNEL FOR HOT UPDATES CLICK HERE TO JOIN OUR TELEGRAM CHANNEL NOW. ENJOY πŸ‘‡ https://t.me/EAGLEEYETEAM ===================================== On How To Get Your Live Question And Answer before Exam. CLICK HERE TO CHAT US πŸ‘‡ http://wa.me/+2348023343546 We Remain The No. 1 West African Examination Official Number 08023343546 ======================================== ========================================

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CHEMISTRY SOLUTION ERN 1A A transmission element is a component that transfers data or energy from one point to another. 1bi. Element D-288 forms a doubly charged cation. 1bii. The soluble trioxocarbonate (IV) is CO32-. 1C. The general decrease in the first ionization energies of the period in the periodic table is due to the increase in atomic radius and the decrease in effective nuclear charge. 1d. -methane (CH4) -propane (C3H8). 1e. Alkanols are stronger bases than water because they have a higher tendency to donate a proton (H+) to an acid. This is because the alkyl group in alkanols is electron-donating, which increases the electron density on the oxygen atom in the hydroxyl group (-OH). 1f. The major raw materials used in the Solvay process are -salt (NaCl) -limestone (CaCO3) -ammonia (NH3). 1g. Geometric isomerism is a type of stereoisomerism that arises when two or more compounds have the same molecular formula and connectivity, but differ in the spatial arrangement of their atoms due to restricted rotation around a double bond or ring. 1h. Water gas is a better fuel than producer gas because it has a higher calorific value and a higher percentage of hydrogen gas. 1hi. Heat of combustion is the amount of heat energy released when one mole of a substance undergoes complete combustion with oxygen under standard conditions. 1ji. Faraday's second law of electrolysis states that the amount of substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. It can be expressed mathematically as: m = (Q * M) / (n * F) 1jii. To calculate the amount of silver deposited, we can use Faraday's law of electrolysis, which states that: m = (Q * M) / (n * F) Plugging in the values, we get: m = (10920 C * 107.87 g/mol) / (1 * 96500 C/mol) m = 12.17 g Therefore, the amount of silver deposited when 10920 coulombs of electricity is passed through a solution of a silver salt is 12.17 grams.

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