Can someone please help me.

Based on your answer in part B, arrange the five elements in order of their reactivity. Drag the most reactive element to the to

amm1812

Answer: Order of the elements from most reactive to least reactive is:

Mg > Zn > Fe > H > Cu

Explanation: Reactivity of the elements is the tendency of the elements to loose or gain electrons with ease.

As, all the given elements are metals, so the metals which will loose electrons easily will be more reactive. To study the reactivity of metals, we use reactivity series.

The metals which are more reactive are placed on the top of the reactivity series and which are less reactive are placed at the bottom.

Hence, the order of the elements from most reactive to least reactive is:

Mg > Zn > Fe > H > Cu

Reactivity series is attached below.

6 0

4 years ago

How do the melting and boiling points of water compare to that of other molecular compounds of similar molar mass?

sergey [27]

Answer:

The answer is D.

Explanation:

3 0

3 years ago

What is the volume of 60 g of ether if the density of ether is 0.70g/mL

miss Akunina [59]

Answer:

114mL.

Explanation: hope this helped

8 0

3 years ago

Please help! The Michael reaction is a conjugate addition reaction between a stable nucleophilic enolate ion (the donor) and an

labwork [276]

Answer:

See explaination andd attachment

Explanation:

Treatment of aldehydes and ketones with a suitable base can lead to the formation of a nucleophilic species called an enolate that reacts with electrophiles. These C nucleophiles are useful for making new carbon-carbon bonds.

Please kindly see attachment for the drawings.

7 0

3 years ago

The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: CO(g) + H2O(g)

taurus [48]

<u>Answer:</u> The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

<u>Explanation:</u>

For the given chemical reaction:

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}$ ........(1)

We are given:

$[CO]_{eq}=[H_2O]_{eq}=[H_2]_{eq}=0.10M$

$[CO_2]_{eq}=0.40M$

Putting values in above equation, we get:

$K_c=\frac{0.40\times 0.10}{010\times 0.10}\\\\K_c=4$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of hydrogen gas = 0.30 mol

Volume of solution = 2.0 L

Putting values in above equation, we get:

$\text{Molarity of }H_2=\frac{0.30mol}{2L}=0.15M$

When hydrogen gas is added, the concentration of product gets increased. But, by Le-Chatelier's principle, the equilibrium will shift in the direction where concentration of product must decrease, which is in the backward direction.

Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M

Now, the equilibrium is shifting to the reactant side. The equation follows:

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial: 0.1 0.1 0.4 0.1

At eqllm: 0.1+x 0.1+x 0.4-x 0.25-x

Putting values in expression 1, we get:

$4=\frac{(0.25-x)(0.4-x)}{(0.1+x)(0.1+x)}\\\\3x^2+1.45x-0.06=0\\\\x=0.038,-0.522$

Neglecting the negative value of 'x'

Calculating the concentrations of the species:

Concentration of carbon dioxide = (0.4 - x) = (0.4 - 0.038) = 0.362 M

Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M

Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Hence, the concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

8 0

4 years ago