Which of the following is the reason that the lemurs shown below share many traits?

Based on the balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O calculate the number of excess reagent units remaining when 52 C2H2 mol

erica [24]

Answer:

20 molecules of oxygen gas remains after the reaction.

Explanation:

$2C_2H_2 + 5O_2\rightarrow 4CO_2 + 2H_2O$

Molecules of ethyne = 52

Molecules of oxygen gas = 150

According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.

Then 52 molecules of ethyne will react with:

$\frac{5}{2}\times 52=130 molecules$ of oxygen gas.

As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.

Remaining molecules of recessive reagent = 150 - 130 = 20

20 molecules of oxygen gas remains after the reaction.

5 0

3 years ago

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lilavasa [31]

1.66 M is the concentration of the chemist's working solution.

<h3>What is molarity?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dilute solution of this reactant.

The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.

We want to know the concentration of this diluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:

$n_1$ = $n_2$ (1)

and we also know that:

n = M x $V_2$

If we replace this expression in (1) we have:

$M_1$ x $V_1$ = $M_2$ x $V_2$

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is $M_2$ :

4.93 x 210 = 620 x $M_2$

$M_2$ = 1.66 M

This is the concentration of the solution prepared.

Learn more about molarity here:

brainly.com/question/19517011

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6 0

2 years ago

Describe how oxidation and reduction involve electrons, change oxidation numbers, and combine in

Sholpan [36]

Answer:

Redox

Explanation:

Reduction is gain of electrons

oxidation is loss of electrons

3 0

3 years ago

How does the A Hreaction relate to the A He of molecules involved in a reaction?

igor_vitrenko [27]

Answer:

B. ΔHreaction = ΔH°f reactants- ΔH°f products

Explanation:

<em>Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction</em>.

Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).

For example, in ΔH°f of H₂O, the equation is:

H₂(g) + 1/2O₂(g) → H₂O(g)

The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).

Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:

NaOH + HCl → H₂O + NaCl. ΔHr

The ΔH°f for each substance in the reaction is:

NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH <em>(1)</em>

HCl: 1/2H₂ + 1/2Cl₂ → HCl <em>(2)</em>

H₂O: H₂ + 1/2O₂ → H₂O <em>(3)</em>

NaCl: Na + 1/2Cl₂ → NaCl <em>(4)</em>

The algebraic sum of (3) + (4) is -(ΔH°f reactants):

H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants

This reaction - {(1)+(2)} ΔH°f products

NaOH + HCl → H₂O + NaCl.

ΔHr = ΔH°f reactants- ΔH°f products

In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

8 0

3 years ago

A mixture of helium, nitrogen and oxygen has a total pressure of 821 mmHg. The partial pressure of helium is 105 mmHg, and the p

skad [1K]

Answer:

Total partial pressure, Pt = 821 mm Hg

Partial pressure of Helium, P1 = 105 mm Hg

Partial pressure of Nitrogen, P2 = 312 mm Hg

Partial pressure of Oxygen, P3 = ? mm Hg

According to Dalton's law of Partial pressures,

Pt = P1 + P2 + P3

So, <u>P3 = 404 mm Hg</u>

4 0

2 years ago