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3 years ago

Of the metals pb,hg,an,and mg, which will not spontaneously donate electrons to copper in solution?

Chemistry

1 answer:

tekilochka [14]3 years ago

5 0

Answer is: Hg (mercury).

<span> Reactivity series is an empirical progression of a series of metals, arranged by their reactivity from highest to lowest (alkaline metals have highest reactivity and Noble metals lowest reactivity). This series are used to summarize information about the reactions of metals with acids or water, double displacement reactions (more reactive metals displace metals with lower reactivity) and the extraction of metals from their ores.</span>

In this example, magnesium and lead are more reactive than copper, mercury is less reactive.

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A student is doing an experiment that stoichiometry says should produce 34.6 grams of product. The student actually makes 25.2 g

attashe74 [19]

25.2 / 34.6 x 100
which would get you 73%

if you found my answer helpful pls give brainliest

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2 years ago

A 5.0 l flask containing o2 at 2.00 atm is connected to a 3.0 l flask containing h2 at 4.00 atm and the gases are allowed to mix

gladu [14]

0.55[that is the correct answer to the question]<<<

6 0

3 years ago

Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/

Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of $C_2H_4$ and $H_2$ = 25.0 atm

Temperature of $C_2H_4$ and $H_2$ = $250^oC=273+250=523K$

Volume of $C_2H_4$ = 1050 L per min

Volume of $H_2$ = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of $C_2H_6$ = 30 g/mole

First we have to calculate the moles of $C_2H_4$ and $H_2$ by using ideal gas equation.

For $C_2H_4$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=611.34moles$

For $H_2$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=902.46moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_2H_4+H_2\rightarrow C_2H_6$

From the balanced reaction we conclude that

As, 1 mole of $C_2H_4$ react with 1 mole of $H_2$

So, 611.34 mole of $C_2H_4$ react with 611.34 mole of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $C_2H_6$ .

As, 1 mole of $C_2H_4$ react to give 1 mole of $C_2H_6$

As, 611.34 mole of $C_2H_4$ react to give 611.34 mole of $C_2H_6$

Now we have to calculate the mass of $C_2H_6$ .

$\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6$

$\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g$

The theoretical yield of $C_2H_6$ = 18340.2 g

The actual yield of $C_2H_6$ = 14.0 kg = 14000 g (1 kg = 1000 g)

Now we have to calculate the percent yield of $C_2H_6$

$\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%$

Therefore, the percent yield of the reaction is, 76.34 %

5 0

3 years ago

How do you determine the percent yield of a chemical reaction?

valentinak56 [21]

Answer:

percentage yield =[ experimental yield / theoretical yield] x 100

Explanation:

7 0

3 years ago

for the mini turbines to work properly, which of the following criteria are important in choosing the material for their constru

Setler79 [48]

Answer:

metal

Explanation:

metal strong

3 0

3 years ago