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Yuki888 [10]
2 years ago
7

3Ba+Al2(SO4)3 -->2Al+3BaSO4,

Chemistry
1 answer:
daser333 [38]2 years ago
6 0

Answer:

13.5 * 10^-2 g

Explanation:

What we know:

Balanced Equation: 3Ba+Al2(SO4)3 -->2Al+3BaSO4,

Grams of Ba: 1

Grams of Al2(SO4)3: 1.8g

Calculate the # of moles of Ba and Al2(SO4)3:

1g Ba/137.3 = 7.3 *10^-3 mol Ba

1.8g Al2(SO4)3/ 342 = 5.3 *10^-3 mol Al2(SO4)3

Find the limiting reactant:

Ba has a coefficient of 3 in the balanced equation, so we divide the # of moles of Ba by 3 to get... 7.3 *10^-3 mol Ba/3 = 2.43 *10^-3

Al2(SO4)3 has a coefficient of 1, so if we divide by 1, we get the same number of 5.3 *10^-3

2.43 *10^-3 is smaller than 5.3 *10^-3, therefore Ba is the limiting reactant.

finally, we just find the number of moles of Al

The ratio of Al to Ba is 2:3 so...

7.3 * 10^-3 * (2/3) = 5 *10^-3 mol Al

CONVERT TO GRAMS

5 *10^-3 mol Al  * 27 = 13.5 * 10^-2 g

<u>Hope that was helpful! </u>

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Answer:

<h3>no it is not allowed</h3>

Explanation:

<h3>Liwis structure shows the elements symbol with dots thet represents valance electrons ; in second row elements their atomic number is 3 up to 10 , from Li up to Ne from their electron configuration their valance electron will be from 1 up to 8 respectivelly ,if lewis structure represents the element with it is symbol and dots that represents valance electron the second row elements cannot have more than an octet of valance electrons surrounding it.</h3>

<h3>I think it is help ful for you </h3>
3 0
3 years ago
. a large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is sub
elena-s [515]

The large piece of jewelry  that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml

To solve this problem the formulas and the procedures that we have to use  are:

  • v = v(f)-v(i)
  • d = m/v

Where:

  • d= density
  • m= mass
  • v= volume
  • v(f) = final volume
  • v(i) = initial volume

Information about the problem:

  • m = 132.6 g
  • v(i) = 48.6 ml
  • v(f) = 61.2 ml
  • v = ?
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Applying the volume formula we get:

v = v(f)-v(i)

v = 61.2 ml - 48.6 ml

v = 12.6 ml

Applying the density  formula we get:

d = m/v

d = 132.6 g/12.6 ml

d = 10.523 g/ml

<h3>What is density?</h3>

It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.

Learn more about density in: brainly.com/question/1354972

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Hydrazine (N 2 H 4 )) a rocket fuel reacts with oxygen to form nitrogen gas and water vapor . The reaction is represented with t
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The mass of hydrazine (N₂H₄) required to produce 96 g of water (H₂O) is 85.4 g (Option C)

<h3>Balanced equation </h3>

N₂H₄ + O₂ —> N₂ + 2H₂O

Molar mass of N₂H₄ = (2×14) + (4×1) = 32 g/mol

Mass of N₂H₄ from the balanced equation = 1 × 32 = 32 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

<h3>How to determine the mass of N₂H₄</h3>

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

Therefore,

96 g of H₂O will be produced by = (96 × 32) / 36 = 85.4 g of N₂H₄

Thus, 85.4 g of N₂H₄ is needed for the reaction

Learn more about stoichiometry:

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the answer i believe is electrolytes.

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