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Nikolay [14]
3 years ago
8

the actual yield of a certain reaction is 38.0 g while the theoretical yield is 50.0 g calculate the percent yield

Chemistry
1 answer:
Verizon [17]3 years ago
8 0

Answer:

76%

Explanation:

Actual yield = 38.0 g

Theoretical yield = 50.0 g

Percentage yield = ?

The relationship between these quantities is given as;

Percent yield = Actual yield / Theoretical yield   *  100%

Percent yield = 38 g / 50 g    *  100 %

Percent yield = 0.76  *  100 % = 76%

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(a) (1)
Elis [28]

Explanation:

The ionization energy of an atom is the amount of energy that is required to remove an electron from a mole of atoms in the gas phase:

M(g)  ®  M+(g)  +  e-

It is possible to remove more electrons from most elements, so this quantity is more precisely known as the first ionization energy, the energy to go from neutral atoms to cations with a 1+ charge.  The second ionization energy is the energy that is required to remove a second electron, to form 2+ cations from 1+ cations:

M+(g)  ®  M2+(g)  +  e-

The third ionization energy is the energy required to form 3+ cations:

M2+(g)  ®  M3+(g)  +  e-

and so on.  Ionization energies are always positive numbers, because energy must be supplied (an endothermic energy change) to separate electrons from atoms.  The second ionization energy is always larger than the first ionization energy, because it requires even more energy to remove an electron from a cation than it is from a neutral atom.

The first ionization energy varies in a predictable way across the periodic table.  The ionization energy decreases from top to bottom in groups, and increases from left to right across a period.  Thus, helium has the largest first ionization energy, while francium has one of the lowest.

From top to bottom in a group, orbitals corresponding to higher values of the principal quantum number (n) are being added, which are on average further away from the nucleus.  Since the outermost electrons are further away, they are less strongly attracted by the nucleus, and are easier to remove, corresponding to a lower value for the first ionization energy.From left to right across a period, more protons are being added to the nucleus, but the number of electrons in the inner, lower-energy shells remains the same.  The valence electrons feel a higher effective nuclear charge — the sum of the charges on the protons in the nucleus and the charges on the inner, core electrons.  The valence electrons are therefore held more tightly, the atom decreases in size (see atomic radius), and it becomes increasingly difficult to remove them, corresponding to a higher value for the first ionization energy.

 

The following charts illustrate the general trends in the first ionization energy:

Dunno kung tama beng pero trysorry kung mali

8 0
3 years ago
A first-order reaction has a half-life of 29.2 s . how long does it take for the concentration of the reactant in the reaction t
Dmitriy789 [7]

Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value

<span> The half-life for the chemical reaction is 29,2 s and is independent of initial concentration.
c</span>₀ - initial concentration the reactant.

c - concentration of the reactant remaining at time.

t = 29,2 s.<span>
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s</span> = 0,0237 1/s.<span>
ln(c/c</span>₀) = -k·t₁.<span>
ln(1/16 </span>÷ 1) = -0,0237 1/s · t₁.

t₁ = 116,8 s.

5 0
4 years ago
If m&lt; CAB = 80 degrees, then find m&lt; BAD
eimsori [14]

Answer:

30

Explanation:

Just trust me

7 0
3 years ago
What are the resulting coefficients when you balance the chemical equation for the combustion of ethane, c2h6? in this reaction,
Vitek1552 [10]

Balance equation for combustion of ethane will be:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

To balance the equation:

1. Balance the number of carbon atom on both side:

C₂H₆(g) + O₂(g)--------> CO₂(g) + H₂O(g)

1. balance the number of carbon on both side, as in reactant there are 2 but in product one,

so , multiply the CO₂, by 2 in the product.

2. Balance the number of hydrogen on both side as in reactant the number of hydrogen is 3 but in product it is 6 so, multiplythe number of  H₂O by 3,

so multiply the number of  H₂O by 3 in product.

3. Balance the number of oxygen on both side , as 1 and 2 step increases the number of oxygen and it becomes 7 , so to balance the number of oxygen on both side by mutiplying the  number of  O₂ by 7/2 in reactant .

4. Now, doubling the equation will give balance equation that is:

2C₂H₆(g) + 7O₂(g)--------> 4CO₂(g) + 6H₂O(g)

8 0
3 years ago
Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (A
UkoKoshka [18]

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ <em>just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>

<em />

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

<h3>[I⁻] = 1.67x10⁻³</h3><h3 />

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>

6 0
3 years ago
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