Calculate the volume of 36 gram of H2O (water) at NTP conditions
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Firstly , IF u write the formula on basis their weights then : 2 * 1 + 1 / 2 * 32 18 1 mole of oxygen = > 18 18 Ratio of oxygen in 18g of water = 16/18 Therefore , amount of Oxygen in 36g of Water will be = 32g Now , mole ( amount of gas ) = Weight in Gram / Molecular weight 32/32 AT STP Volume of 1 mole of any gas is 22.4 L will be required . Therefore , 22.4 L Oxygen is required . BUDDY PLZ MARK IT BRAINLIEST IF U FEEL IT DESERVES
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