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lys-0071 [83]
3 years ago
8

the experimental yield of a compound is 30 g and the true yield is 40 g what is the percentage error in calculating the yield

Chemistry
1 answer:
AleksAgata [21]3 years ago
3 0

Answer:

25%.

Explanation:

The following data were obtained from the question:

Experimental yield = 30 g

Actual yield = 40 g

Percentage error =..?

Thus, we can calculate the percentage error as follow:

Percentage error = |Experimental – Actual | / Actual × 100

Percentage error = |30 – 40|/40 × 100

Percentage error = 10/40 × 100

Percentage error = 25 %

Therefore, the percentage error in calculating the yield is 25%

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Collision theory states that as molecules or ions bump into each other, a reaction will only occur if the collision has the corr
Vanyuwa [196]

Answer: As temperature increases, the number of collisions increases and the energy of the collisions increases.

Explanation:

According to collision theory, for a reaction to take place it is necessary to have collisions between the reacting species or atoms.

A collision will only be effective if species coming together have a certain minimum value of internal energy equal to the activation energy of the reaction.

More is the number of collisions taking place in a chemical reaction more will be the kinetic energy of its molecules. As kinetic energy is the energy acquired due to motion of atoms or a substance.

Also, collisions increases with increase in temperature as:

K.E = \frac{3}{2}kT

Kinetic energy is directly proportional to temperature. So, more is the temperature more will be energy of molecules.

Thus, we can conclude that as temperature increases, the number of collisions increases and the energy of the collisions increases.

7 0
3 years ago
The equilibrium constant for the reaction Ni2+(aq) + 6 NH3(aq) ⇌ Ni(NH3)6 2+(aq) is Kf = 5.6 × 108 at 25°C. (a) What is ΔG o at
shusha [124]

Answer:

(a) -49.9 kJ/mol;

(b) To the right;

(c) 34.6 kJ/mol

Explanation:

(a) For this reaction, since it's at equilibrium and standard states, we know that we can apply the equation:

\Delta G^o = -RT ln (K_f)

Substituting the given variables:

\Delta G^o = -8.314 \frac{J}{K mol}\cdot 298.15 K\cdot ln (5.6\cdot 10^8) = -49932 J/mol = -49.9 kJ/mol

(b) Notice that this reaction is spontaneous, since \Delta G^o < 0. This means reaction spontaneously proceeds to the right side. Besides, K > 1, this means products dominate over reactants, so reaction proceeds to the right.

(c) Given the expression of the formation constant, we can use the same expression to calculate the reaction quotient at non-standard conditions:

Q_f = \frac{[Ni(NH_3)_6]^{2+}}{[Ni^{2+}][NH_3]^6} = \frac{0.010}{0.0010\cdot 0.0050^6} = 6.4\cdot 10^{14}

Now, notice that Q_f > K_f. In this case, we have an excess of the products, this means reaction will shift to the let left to restore the equilibrium.

Calculate:

\Delta G = \Delta G^o + RT ln Q_f = -49932 J/mol + 8.314 \frac{J}{K mol}\cdot 298.15 K\cdot ln(6.4\cdot 10^{14}) = 34577 J = 34.6 kJ/mol

5 0
3 years ago
What is the total number of carbon atoms in a molecule of ethanoic acid?
Marrrta [24]
<span>There are 2 carbon atoms in ethanoic acid. Other name of such substance is acetic acid. It is a colorless liquid carboxylic acid with the chemical formula CH3COOH. It has antibacterial and antifungal properties.</span>
8 0
2 years ago
How many moles of helium are 8.84×10^24 atoms of He?
viva [34]

Answer:

14.77 mol.

Explanation:

  • It is known that every 1.0 mole of compound or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

<u><em>Using cross multiplication:</em></u>

1.0 mole of He contains → 6.022 x 10²³ atoms.

??? mole of He contains → 8.84 x 10²⁴ atoms.

<em>∴ The no. of moles of He contains (8.84 x 10²⁴ atoms) </em>= (1.0 mol)(8.84 x 10²⁴ atoms)/(6.022 x 10²³ atoms) =<em> 14.77 mol.</em>

8 0
3 years ago
A chemist fills a reaction vessel with mercurous chloride solid, mercury (I) aqueous solution, and chloride aqueous solution at
makvit [3.9K]

Answer:

ΔG° = -533.64 kJ

Explanation:

Let's consider the following reaction.

Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)

The standard Gibbs free energy (ΔG°) can be calculated using the following expression:

ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)

where,

ni are the moles of reactants and products

ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products

ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)

ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)

ΔG° = -533.64 kJ

3 0
3 years ago
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