Use the periodic table to complete each nuclear fusion equation.

alex41 [277]

Answer:

Percent Yield Fe = 82.5%

Explanation:

The actual yield is the value produced after an experiment is conducted. The theoretical yield is the value calculated using the balanced chemical equation and atomic/molar masses.

To find the percent yield of iron (Fe), you need to (1) convert grams Al to moles Al (via atomic mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from equation coefficients), then (3) convert moles Fe to grams Fe (via atomic mass), and then (4) calculate the percent yield. It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the sig figs of the given values.

Atomic Mass (Mg): 24.305 g/mol

Atomic Mass (Fe): 55.845 g/mol

3 Mg + 2 FeCl₃ -----> 2 Fe + 3 MgCl₂

20.5 g Mg 1 mole 2 moles Fe 55.845 g
----------------- x ----------------- x ---------------------- x ----------------- =
24.305 g 3 moles Mg 1 mole

= 31.4 g Fe

Actual Yield
Percent Yield = ---------------------------------- x 100%
Theoretical Yield

25.9 g Fe
Percent Yield = -------------------- x 100%
31.4 g Fe

Percent Yield = 82.5%

5 0

2 years ago

In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in

Ilia_Sergeevich [38]

Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

[BF4-] = 0.1326 M

From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

Forward rate = 9.00 × 10^(-4) /min × (0.1326M)

Forward rate = 1.1934 × 10^(-4) M/min

(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

Thus, I disagree with the claim.

3 0

3 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

3 years ago

Conserving water can save money while protecting the environment. True or False

blagie [28]

Answer: The given statement is true.

Explanation:

Water is a natural resource present in the nature and it is very precious as life without it is impossible.

So, when we unnecessarily use water then it means we are wasting a natural resource that cannot be reproduced again by human beings.

Therefore, more we are able to conserve water more we can protect other human life's and environment also.

If there is shortage of water then its prices will go high and hence we need to pay more for it.

Thus, we can conclude that the statement conserving water can save money while protecting the environment, is true.

3 0

3 years ago

Read 2 more answers

When magnesium metal is heated in air it begins to release large amounts of heat and light. What kind of a reaction is this an e

Salsk061 [2.6K]

It is called exothermic reaction because it releases heat and light and it is called combustion reaction because it is reacting and is being oxidised by O2 to MgO. It can also be called as oxidation reaction since Mg is oxidised to MgO.

5 0

3 years ago

Read 2 more answers