Answer:
4.78 %.
Explanation:
<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>
<em></em>
<em>mass % = (mass of solute/mass of solution) x 100.</em>
<em></em>
mass of MgSO₄ = 50.0 g,
mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.
mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.
<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
The temperature will change from 100K to 173.87 K
calculation
by use of law that is V1/T1=V2/T2
V1=3.75 L
T1=100k
V2=6.53 L
T2=?
make T2 the subject of the formula
T2=(V2 xT1)V1
=6.52 x100/3.75=173.87K
Answer:
The number on the lag label should be 15.
Explanation:
It seems your question is incomplete, as it is lacking the working values. An internet search showed me the full question, you can see it in the attached picture.
Let's say we have 100 g of the fertilizer.
- <em>45 g are of ammonium phosphate</em> ( (NH₄)₃PO₄ ), of which:
- 45 g (NH₄)₃PO₄ *
= 12.7 g are of Nitrogen.
(We used the molar mass of ammonium phosphate in the denominator and three times the molar mass of nitrogen in the numerator)
- <em>18 g are of calcium nitrate</em> (Ca(NO₃)₂), of which:
- 16 g Ca(NO₃)₂ *
= 2.73 g are of Nitrogen.
So in total there are (12.7+2.73) 15.43 g of Nitrogen in 100 g of the fertilizer. So the percent by mass of nitrogen is 15.43%.
Rounding to the nearest percent the answer is 15.
Answer: Option (b) is the correct answer.
Explanation:
It is given that mass of Mg is 97.22 g and it is known that molar mass of Mg is 24.305 g/mol.
So, calculate the number of moles as follows.
No. of moles = 
=
= 4 mol
Also, it is known that 1 mole has
atoms/mol. Therefore, calculate the number of atoms in 4 mol as follows.

=
atoms
or, =
atoms
Thus, we can conclude that there are
atoms in 97.22 grams of Mg.