Answer:
a. the mole fraction of CO in the mixture of CO and O2.
mole fraction = moles of CO/ Total moles of the mixture
Mole fraction of CO = 10/(10+12.5)=0.444
b. Reaction - CO(g)+½O2(g)→CO2(g)
Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2
So given,
At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.
3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2
This means that unused mols are : 7mols of CO and 11mols of O2
Total product mixture = 3 + 7 + 11 = 21mols
mole fraction of CO = 7/21 = 0.33
Answer:
2HNO3 + Ba(OH)2 => Ba(NO3)2 + 2H2O
Explanation:
No they wouldn't. <span>You can't make an </span>ionic compound<span> with these elements.</span>
When `CO_(2)` is bubbled through a cold pasty solution of barium peroxide in water, `H_(2)O_(2)` is obtained. <br> `BaO+CO_(2)+H_(2)OtoBaCO_(3)+H_(2)O_(2)` Barium carbonate being insoluble is filtered off. This is known as Merck's process.
<h3>What is meant by Perhydrol?</h3>
perhydrol (countable and uncountable, plural perhydrols) A stabilised solution of hydrogen peroxide.
<h3>What is Merck's Perhydrol?</h3>
Uses: Perhydrol is used as an antiseptic for wounds, and also acts as a germicide to kill bacteria and germs.
Being a strong oxidizing agent it has bleaching properties and acts as a ripening agent.
Learn more about merck's process here:
<h3>
brainly.com/question/16856280</h3><h3 /><h3>#SPJ4</h3>
Answer:
Explanation:
1. Moles of CCl₄
2. Molar mass of CCl₄
MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol
3. Mass of CCl₄
4. Volume of CCl₄
