The reaction for the formation of MgO(s):
2 Mg (s) + O2(g) -à
2MgO(s) ΔH = -601.24
kJ/mol
<span>The enthalpy
information is taken from: http://webbook.nist.gov/cgi/inchi?ID=C1309484&Mask=2</span>
From the equation and with an enthalpy change of -231 kJ:
-231 kJ * 2 mol Mg * (1/-601.24 kJ/mol) = 0.76841 mol Mg
Then, with the molar mass of MgO = 40.3,
0.76841 mol Mg *(2 mol MgO/2 mol Mg)* 40.3 g/mol MgO = <span>30.967 g MgO</span>
Answer:
1.67g H2CO3 are produced
Explanation:
Based on the reaction:
2NaHCO3 → Na2CO3 + H2CO3
<em>2 moles of NaHCO3 produce 1 mole of Na2CO3 and 1 mole of H2CO3</em>
To solve this question we need to find the moles of Na2CO3 = Moles of H2CO3. With their moles we can find the mass of H2CO3 as follows:
<em>Moles Na2CO3 -Molar mass: 105.99g/mol-</em>
2.86g Na2CO3 * (1mol/105.99g) = 0.02698 moles Na2CO3 = Moles H2CO3
<em>Mass H2CO3 -Molar mass: 62.03g/mol-</em>
0.02698 moles * (62.03g/mol) =
<h3>1.67g H2CO3 are produced</h3>
Carbon dioxide pass through a leaf stomata.
Answer:
A mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly
12 grams of carbon-12 (i.e., 6.023 X 1023).
Explanation:
