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2 years ago

What is the percent yield if the theoretical yield of ChCl3 is 14.50 g but the actual yield is 12.33 g?

1 answer:

4 0

Percentage is a ration of part to a whole. In our case, in perfect conditions, yield is theoretical and thusly considered larger, a whole. While actual yield is less because of imperfect conditions. The percentage is,

$(12.33\cdot 100)/14.5\approx85.03\%$ .

Hope this helps.

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Pani-rosa [81]

Answer: Do want the experiment done or help of the experiment

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2 years ago

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A block is pulled 0.90 m to the right in 3.75 s.

VLD [36.1K]

Answer:

0.36 m/s

Explanation:

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2 years ago

Please help!!!!

Nat2105 [25]

Answer:

When an atom gains or loses energy, the energy of an electron can change.

An electron in an atom can move from one energy level to another when the atom gains or loses energy.

Explanation:

The possible energies that electrons in an atom

can have are called energy levels.

• An electron cannot exist between energy levels.

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2 years ago

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Calcium nitrate and ammonium fluoride react to form calcium fluoride, dinitrogen monoxide, and water vapor. How many grams of di

pychu [463]

Answer:

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:

$Ca(NO_3)_2(aq)+2NH_4F(aq)\rightarrow CaF_2(aq)+2N_2O(g)+4H_2O(g)$

Moles of calcium nitrate = $\frac{31.3 g}{164 g/mol}=0.1908 mol$

Moles of ammonium fluoride = $\frac{38.7 g}{37 g/mol}=1.046 mol$

According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.

Then 1.046 moles of ammonium fluoride will react with :

$\frac{1}{2}\times 1.046 mol=0.523 mol$ calcium nitarte .

This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.

Hence, calcium nitrate is a limiting reactant.

So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.

So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .

Then 0.1908 moles of calcium nitrate will give:

$\frac{2}{1}\times 0.1908 mol=0.3816 mol$ of dinitrogen monoxide gas.

Mass of 0.03816 moles of dinitrogen monoxide gas:

0.03816 mol × 44 g/mol = 16.79 g

16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.

8 0

2 years ago

Which is least likely to be reduced? <br> A. Zn^2+ <br> B. Fe^3+ <br> C. Cu^2+ <br> D. Fe^2+

alekssr [168]

A. Zn²⁺

<h3>Further explanation</h3>

Given

Cations of several elements

Required

The least to be reduced

Solution

If we look at the voltaic series:

The electrode which is easier to reduce than the hydrogen (H2) electrode has a positive sign (E red= +) and is located to the right of the voltaic series (right of H)

The electrode which is easier to oxidize than the hydrogen (H2) electrode and is difficult to experience reduction has a negative sign (E red= -) and is located to the left of the voltaic series (left of H)

Or you can look at the standard reduction potential value of the metals in the answer options, and the most negative reduction E° value which will be difficult to reduce.

The Zn metal is located far left of the other metals in the answer choices, so it is the most difficult to reduce

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2 years ago