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hjlf
3 years ago
6

When mixed together, baking soda and vinegar undergo a chemical reaction. Carbon dioxide, water, and

Chemistry
2 answers:
Valentin [98]3 years ago
8 0

Answer: baking soda + vinegar →  carbon dioxide + water + sodium acetate

Simora [160]3 years ago
7 0

Answer:

Baking soda + vinegar------>Carbon dioxide + sodium acetate+ water

Explanation:

Vinegar is obtained from the oxidation of ethanol in wines to ethanoic acid in the air. The CO2 produced could be used to inflate balloons.

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How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield
ahrayia [7]
<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>

3 0
3 years ago
Read 2 more answers
Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β
Fantom [35]
<h3>Answer:</h3>

8 alpha particles

4 beta particles

<h3>Explanation:</h3>

<u>We are given;</u>

  • Neptunium-237
  • Thallium-205
  • Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

We are required to determine the number of beta and alpha particles produced to complete the decay series.

  • We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
  • When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

In this case;

Neptunium-237 has an atomic number 93, while,

Thallium-205 has an atomic number 81.

Therefore;

²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

We can get x and y

237 = 4x + y(0) + 205

237-205 = 4x

4x = 32

 x = 8

On the other hand;

93 = 2x + (-y) + 81

but x = 8

93 = 16 -y + 81

y = 4

Therefore, the complete decay equation is;

²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.

5 0
2 years ago
If element A is in Group 2 and element B is in Group 17, what will be the formula if they bond together?
Anit [1.1K]

Answer:

AB

Explanation:

When combining two elements in different groups, you would just put them together unless you have two or more of each element

4 0
2 years ago
If you were to descend through the ocean in a submersible, what changes in the ocean water could you observe?
aalyn [17]

Answer:

it would become darker. there would be a higher level of water pressure. and there would be less animals, or different alimals.

5 0
3 years ago
Number of grams of hydrogen than can be prepared from 6.80g of aluminum​
Anastaziya [24]

Answer:

0.7561 g.

Explanation:

  • The hydrogen than can be prepared from Al according to the balanced equation:

<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>

It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.

  • Firstly, we need to calculate the no. of moles of (6.8 g) of Al:

no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.

<em>Using cross multiplication:</em>

2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.

0.252 mol of Al need to react → ??? mol of H₂.

∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.

  • Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:

mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.

5 0
3 years ago
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