It’s a.carbon according to Siri
Mass of BaO in initial mixture = 3.50g
Explanation:
Let mass of BaO in mixture be x g
mass of MgO in mixture be (6.35 - x) g
Initially CO_2
Volume = 3.50 L
Temp = 303 K
Pressure = 750 torr = 750 / 760 atm
Applying ideal gas equation
PV = nRT
n = PV / RT
(n)_CO_2 = ((750/760)* 3.50) / 0.0821 * 303
(n)_CO_2 = 0.139 mole
Finally; mole of CO_2
n= PV /RT
((245/760) *3.5) / 303* 0.0821
(n)_CO_2 = 0.045 mole
Mole of CO_2 reacted = 0.139 - 0.045
=0.044 mole
BaO + CO_2 BaCO_3
Mgo + CO_2 MgCO_3
moles of CO_2 reacted = ( moles of BaO + moles of MgO)
moles of BaO in mixture = x / 153 mole
moles of MgO in mixture = 6.35 - x mole / 40
Equating,
x/ 153 +6.35/40 = 0.094
= x/153 + 6.35 / 40 - x/40 =0.094
= x (1/40 - 1153) = (6.35/40 - 0.094)
= x * 10.018464
= 0.06475
mass of BaO in mixture = 3.50g
Answer:
12.9 g O₂
Explanation:
To find the mass of oxygen gas produced, you need to (1) convert grams KClO₃ to moles KClO₃ (via molar mass from periodic table values), then (2) convert moles KClO₃ to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to match the given value (33.0 g).
Molar Mass (KClO₃): 39.098 g/mol + 35.45 g/mol + 3(15.998 g/mol)
Molar Mass (KClO₃): 122.542 g/mol
2 KClO₃ ---> 2 KCl + 3 O₂
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
33.0 g KClO₃ 1 mole 3 moles O₂ 31.996 g
-------------------- x ------------------- x ----------------------- x ------------------ =
122.542 g 2 moles KClO₃ 1 mole
= 12.9 g O₂
Our blood have a pH around the 7.0, when we hold the breath our cells still breathing, and the release

, this compound react with water (hydrolysis).

This compound formed by this reaction is acid, carbonic acid.
Them the pH will get lower them normal, and our buld send an action to our body, and this reaction will be the <span>breathing.</span>