Answer:
Mg(NO4)2 is 180.3 g/mol
Explanation:
First find the substance formula.
Magnesium Nitrate.
Magnesium is a +2 charge.
Nitrate is a -1 charge.
So to balance the chemical formula,
We need 1 magnesium atom for every nitrate atom.
2(1) + 1(-2) = 0
So the substance formula is Mg(NO4)2.
Now find the molar mass of Mg(NO4)2.
Mg = 24.3 amu
N = 14.0 amu
O = 16.0 amu
They are three nitrogen and twelve oxygen atoms.
So you do this: 24.3 + 14.0(2) + 16.0(8) = 180.3 g/mol
So the molar is mass is 180.3 g/mol.
The final answer is Mg(NO4)2 is 180.3 g/mol
Hope it helped!
<span>B)<span>C2H6O<span>2
</span></span></span>
First, convert each percentage to grams: 38.7g, 9.70g, and 51.6g.
Next, calculate the number of moles of each element, based on the number of grams given.
C = 3.23 mol
H = 8.91 mol
O = 3.23 mol
Set up the ratio of moles of each element:
C3.34H9.70O3.23. Convert the decimals to whole numbers by dividing by the smallest subscript, 3.23.
The empirical formula is CH3O.
Now, compute the formula mass, which is 31. Finally, divide the molecular mass by the formula mass, 62/31 = 2. Multiple the subscripts by 2 to get the molecular formula.
The cations has positive charges that are metals while the anions have negative charges that are non-metals. Upon reaction, there is an exchange in charges that are reflected in the subscripts of the atoms. In this case, compound AX2 must have a cation, A belonging to group 2 A with +2 charge and anion, X belonging to Group 7A with -1 charge. Answer is D.
<span>Answer: 100 ml
</span>
<span>Explanation:
1) Convert 1.38 g of Fe₂S₃ into number of moles, n
</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>
iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
</span>
<span>2) Use the percent yield to calculate the theoretical amount:
</span>
<span>65% = 0.65 = actual yield/ theoretical yield =>
</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>
3) Chemical equation:
</span>
<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)
4) Stoichiometrical mole ratios:
</span>
<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl
5) Proportionality:
</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃
6) convert 0.020 mol to volume
</span>
<span>i) Molarity formula: M = n / V
</span>
<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
</span>
