The answer is 33.33 %
The explanation:
According to the reaction equation:
MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)
we can see that 1 mole of MCO3 will produce → 1 mole of CO2
-Now we need o get number of mole of CO2:
and when we have 0.22 g of CO2, so number of mole = mass / molar mass
moles = 0.22 g / 44 g/mol = 0.005 mole
∴ moles of Mg = moles of CO2 = 0.005 mole
∴ mass of Mg = moles * molar mass
= 0.005 * 84 /mol = 0.42 g
∴ Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100
= 33.33 %
Answer:
Having difficulties !!! screen shotted it :)
16. H20 - Covalent
17. Mn(NO2)2 - Ionic and called Manganese(II) nitrate
18. HgO - Ionic
19. Li3N - Covalent
At the half equivalence point [HA] = [A-] and pH = pKa
<span>if Ka is 5.2e-5 then pKa = pH = 4.28</span>
Assuming that the reactants are:
(NH4)2SO4 (aq) + Ba(NO3)2 (aq)
and the products are:
BaSO4 (s) + 2NH4NO3 (aq),
then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.
According to the solubility rules, the following elements are considered insoluble when paired with SO4:
Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+
Therefore, the precipitate will be BaSO4 (s).
The volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles. Details about volume can be found below.
<h3>How to calculate volume?</h3>
The volume of a gas can be calculated using the following formula:
PV = nRT
- P = pressure
- V = volume
- n = number of moles
- R = gas law constant
- T = temperature
0.75 × 0.652 = n × 0.0821 × 313
0.489 = 25.69n
n = 0.489/25.69
n = 0.019moles
Therefore, the volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles.
Learn more about volume at: brainly.com/question/1578538
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