What is the precipitate for CuSO4+NaOH<br> (Balancing Equations)

A student creates a model that displays a cross-section of a metal wire wrapped in plastic insulation. Which statement best desc

artcher [175]

Cross-section of a metal wire wrapped in plastic insulation the metal wire is an element.

<h3>What is an element?</h3>

An element is a substance or material that cannot be affected by an external substance or cannot be broken down by any reaction.

Metal wire is serving as an element, it should not be easily broken or penetrated as an insulator.

Therefore, Cross-section of a metal wire wrapped in plastic insulation the metal wire is an element.

Learn more on element here,

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7 0

2 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

Please someone help mee

Nikitich [7]

Answer:

13.9kj/mol

Explanation:

$h = 891 -(393.5 +2*241.8)=1768.1=13.9kj/mol$

8 0

3 years ago

Refer to the following illustration. How many hydrogen atoms could bond with oxygen?

WITCHER [35]

2.

The reason why is because there are 2 available electron spots on the orbital for the oxygen atom. Hydrogen atoms have one proton and one electron, thus, in order to fill the oxygen atom orbital to a full outer shell, a maximum of 2 atoms could bond with the oxygen atoms.

6 electrons (oxygen)+ 1 electron (hydrogen)+ 1 electron (hydrogen)= 8

4 0

3 years ago

Read 2 more answers

An aluminum atom has a mass of 4.48 * 10-23 g and a small airplane has a mass of 5000 kg. Use this information to answer the que

Rudiy27

Answer:

a. 27g/mol

b. 1.85 x 10^5 moles

Explanation:Please see attachment for explanation

4 0

3 years ago

What is the precipitate for CuSO4+NaOH (Balancing Equations)