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svetlana [45]
2 years ago
11

if m&m are added to the periodic table as a "new" element what would their atomic mass be? a standard bag of m&ms = 1.75

oz and contains about 55 m&ms. (16oz = 453.6 g) put answer in (g/mol)​
Chemistry
1 answer:
Allushta [10]2 years ago
4 0

Answer:  5.43x10^23 g/mole M&M's

Explanation:

1 bag of M&M's = 1.75oz       (1.75oz)*(453.6g/16oz) = 49.61 g

1 bag = 55 M&M's

(49.61 g)/55 M&M = 0.9021 g/MM

1 mole M&M's = 6.023x10^23 M&M's

(6.023x10^23 M&M's)*(0.9021 g/MM's) = 5.43x10^23 g/mole  molar mass

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A current is applied to two electrolytic cells in series. In the first, silver is deposited; in the second, a zinc electrode is
vitfil [10]

The amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .

Given ,

A current is applied to two electrolytic cells in series .

In the first ,silver is deposited and in the second a zinc electrode is consumed .

the reactions involving are ;

Ag+ (aq) + e = Ag

Zn = Zn2+ (aq) +2e

thus the resultant equation is ,

2Ag+ (aq) +Zn = 2Ag + Zn2+

Thus for every mole of Zn dissolves , there is 2 moles of Ag is formed .

65.38 g of Zn contains = 1 moles

1.2 g of Zn contains = 1.2/65.38 =0.01835 moles

for every 1 mole of Zn dissolves there is 2 moles of Ag formed .

Thus the amount of Ag formed in moles =2(O.01835) =0.0367 Moles

1 mole of Ag contains = 107.86g

0.0367 moles of Ag contains = 107.86 (0.0367) =3.959 g of Ag

Hence ,the amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .

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6 0
2 years ago
32. Why do you think the Cl-ion is larger than a neutral Cl atom?
Marat540 [252]
because of ur mom lol
5 0
3 years ago
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How many milliliter of a solution of 4.00KI are needed to prepare 250.0mL of 0.760 KI
Alexeev081 [22]
Answer:

47.5 mL

Solving:

M1 = 4.00 M

V1 = ?

M2 = 0.760 M

V2 = 0.250 L

---

M1 * V1 = M2 * V2

V1 = ( M2 * V2 ) / M1

V1 = ( 0.760 * 0.250 ) / 4.00

V1 = ( 0.190 ) / 4.00

V1 = 0.0475 L
3 0
4 years ago
Q15. Natural gas burns in air to form carbon dioxide and water to releasing heat. CHA(g) O2 (g) CO2 g) H20 (g) AH 802.3 kJ What
mestny [16]

Answer:

a) 0.115 g

Explanation:

The balanced reaction is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

To heat 55g of water, the energy in form of heat necessary can be calculated by:

Q = mcΔT

where Q is the heat, m is the mass, c is the specific heat (for water, c = 4.18 J/gºC), and ΔT is the variation of the temperature, which is 25ºC, so:

Q = 55x4.18x25

Q = 5747.5 J = 5.7475 kJ

So, for the reaction, 1 mol of CH₄ releases 802.3 kJ, so to release 5.7475 kJ will be necessary:

1 mol ---------------- 802.3 kJ

x ---------------- 5.7475 kJ

By a simple direct three rule:

802.3x = 5.7475

x = 7.164x10⁻³mol

The molar mass of CH₄ is : 12 (of C) + 4x1 (of H) = 16 g/mol

The mass is equal to the number of moles multiplied by molar mass, the:

m = 7.164x10⁻³x16

m = 0.115 g

6 0
3 years ago
How many moles of PC15 can be produced from 58.0 g of Cl₂ (and excess<br> P4)?
ludmilkaskok [199]

0.3268 moles of PC15 can be produced from 58.0 g of Cl₂ (and excess

P4)

<h3>How to calculate moles?</h3>

The balanced chemical equation is

P_{4}  + 10Cl_{2}  = 4PCl_{5}

The mass of clorine is m(Cl_{2}) = 58.0 g

The amount of clorine is n(Cl_{2}) = m(Cl_{2})/M(Cl_{2}) = 58/70.906 = 0.817 mol

The stoichiometric reaction,shows that

10 moles of Cl_{2} yield 4 moles of PCl_{5};

0.817 of Cl_{2} yield x moles of PCl_{5}

n(PCl_{5}) = 4*0.817/10 = 0.3268 mol

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3 0
2 years ago
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