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Daniel [21]
2 years ago
5

The list shows the results of six athletes competing in the long jump, which has a target distance of 20 feet.

Chemistry
1 answer:
kati45 [8]2 years ago
7 0

Answer:

Fairly accurate but not precise

Good luck! I hope this helps! <33

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A solution was prepared by mixing 50.0 g
frez [133]

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>

4 0
2 years ago
Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80
AVprozaik [17]

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

4 0
2 years ago
when the volume of a gas is changed from 3.75 L to 6.52 L the temperature will change from 100k to _K
harkovskaia [24]

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


4 0
2 years ago
Read 2 more answers
To help farmers and gardeners, commercial fertilizers have a big three-number "NPK" label on the bag that gives the amounts of t
Ludmilka [50]

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 * \frac{42 g N}{149g(NH_{4})_{3}PO_{4}} = 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 *\frac{28gN}{164gCa(NO_{3})_{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.

8 0
1 year ago
How many atoms of Mg are present in 97.22 grams of Mg?
Lostsunrise [7]

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 = \frac{mass given in grams}{Molar mass}

                                 = \frac{97.22 g}{24.305 g/mol}  

                                 = 4 mol

Also, it is known that 1 mole has 6.023 \times 10^{23} atoms/mol. Therefore, calculate the number of atoms in 4 mol as follows.

               4 mol \times 6.023 \times 10^{23} atoms/mol

               = 24.08 \times 10^{23} atoms

or,            = 2.408 \times 10^{23} atoms

Thus, we can conclude that there are 2.408 \times 10^{23} atoms in 97.22 grams of Mg.

7 0
2 years ago
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