No it will be a 10% of that balance so take 16 * 10% and then take that answer and divide bye 70%
Answer:
24.4 amu or g/mole
Explanation:
24 x 0.790 = 19.0 amu
25 x 0.100 = 2.50 amu
26 x 0.110 = 2.86 amu
(Because of the 19.0, the sig figs go only to the 1/10 decimal place)
19.0 + 2.5 + 2.9 = 24.4 amu or g/mole
Answer:
2.9 grams.
Explanation:
- From the balanced reaction:
<em>Mg + 1/2O₂ → MgO,</em>
1.0 mole of Mg reacts with 0.5 mole of oxygen to produce 1.0 mole of MgO.
- We need to calculate the no. of moles of (1.8 g) of Mg and (6.0 g) of oxygen:
no. of moles of Mg = mass/molar mass = (1.8 g)/(24.3 g/mol) = 0.074 mol.
no. of moles of O₂ = mass/molar mass = (6.0 g)/(16.0 g/mol) = 0.375 mol.
<em>So. 0.074 mol of Mg reacts completely with (0.074/2 = 0.037 mol) of O₂ which be in excess.</em>
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<em><u>Using cross multiplication:</u></em>
1.0 mole of Mg produce → 1.0 mol of MgO.
∴ 0.074 mol of Mg produce → 0.074 mol of MgO.
<em>∴ The amount of MgO produced = no. of moles x molar mass </em>= (0.074 mol)(40.3 g/mol) = <em>2.98 g.</em>
Answer:
Balancing chemical equation means making a number of atoms or molecules equal on both sides. In other words, this means that the number of atoms and molecules of each reacting element needs to be the same as the number of atoms and molecules of those elements in the product.
Our reaction is:
AlBr3 + K2SO4 -> KBr + Al2(SO4)3
and we need to balance it.
Since there are 3 molecules of SO4 in the product we need to put 3 before the reactant K2SO4. There are also 2 atoms of Al in the product, so we need to put 2 in front AlBr3. Now we have 6 atoms of K and Br on the left side, so we need to put 6 in front of KBr in the product.
So, our balanced equation will look like this:
2AlBr3 + 3K2SO4 -> 6KBr + Al2(SO4)3
