Reactant + reactant = product
C + O2 = CO2
carbon dioxide is what's being made is product
Answer:
Explanation:
SnSe2 = Tin (IV) selenide
GaAs = Gallium arsenide
Pb(SO4)2 = Lead(II) sulfate
Be(HCO3)2 =Beryllium Bicarbonate.
Mn2(SO3)3 = Manganese(III) Sulfite.
Al(CN)3 =Aluminum Cyanide
hope it helps :)
Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
<span>0.228 g
The balance formula for the reaction between SrH2 and H2O is
SrH2(s) + 2 H2O(l) ==> Sr(OH)2(s) + 2 H2(g)
So for every mole of SrH2 used, 2 moles of hydrogen gas, or 4 moles of hydrogen atoms are released. So let's calculate the molar mass of SrH2 and H2O so see what the limiting reactant is.
strontium = 87.62
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of SrH2 = 87.62 + 2 * 1.00794 = 89.63588 g/mol
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol
Moles of SrH2 = 5.06 g / 89.63588 g/mol = 0.056450609 mol
Moles of H2O = 4.34 g / 18.01488 g/mol = 0.240911957 mol
Looking at the balanced formula, for every mole of SrH2, it takes 2 moles of H2O. So the limiting reactant will be the SrH2. And for every mole of SrH2 used, we get 4 moles of hydrogen atoms. So
4 * 0.056450609 mole * 1.00794 g/mole = 0.227595307 g
Since we only have 3 significant figures, round the result to 3 figures, giving
0.228 g</span>
<span>Because
the early universe was so hot that even tough the conditions to
create heavy elements were there, they will break apart
immediately after their
formation because it was
too hot for the heavy elements to remain stable. It was only later
after the universe cold down that elements weren’t break apart and
the process of forming heavy elements could begin.</span>
