The correct balanced chemical equation which occur when methane burn in oxygen gas is
CH4 + 2O2 --> 2H2O +CO2 (answer D)
when methane is burned in presence of oxygen gas it produces carbon(IV) oxide and water and energy,that is 1 mole of CH4 reacted with 2 moles of O2 to form 2 moles of water and 1 mole of CO2 and enegy. This energy make the methane to be used as fuel.
Answer:
Explanation:
the molecular mass of Na2C6H6O7 =236 g\mole it has a sodium that has 23 g/mole so 7.6 g of Na2C6H6O7 has x of sodium mass
236 g/mole ⇒ 23g/mole
<h2> 7.6 g ⇒ ˣ </h2>
7.6 x 23 ÷ 236 = 74.07×10-2 grams of sodium
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I believe it is the Medial Temperal Lobe. It is part of the brain<span> known as the limbic system, which includes the hippocampus, the amygdala, the cingulate gyrus, the thalamus, the hypothalamus, the epithalamus, the mammillary body and other organs, many of which are of particular relevance to the processing of </span>memory<span>.</span>
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
