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3 years ago

WHAT DOES THIS SAY AND WHAT THE ANSWER

Chemistry

1 answer:

notka56 [123]3 years ago

4 0

It says

The majority of elements on the periodic table are ______________ (metals, nonmetals, or metalloids).

The periodic table on the left separates elements into three groups: the metals (green in the table), nonmetals (orange), and metalloids (blue). Most elements are metals.

Apr 18, 2003

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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

A chemist prepares a solution of pottasium bromide KBr by measuring out 224.g of pottasium bromide into a 300.mL volumetric flas

avanturin [10]

Explanation:

Below is an attachment containing the solution.

8 0

3 years ago

Read 2 more answers

Refer to the periodic table tool and write the electron configurations of the following elements in both long and short terms

ahrayia [7]

Answer:-

Carbon

[He] 2s2 2p2

1s2 2s2 2p2.

potassium

[Ar] 4s1.

1s2 2s2 2p6 3s2 3p6 4s1

Explanation:-

For writing the short form of the electronic configuration we look for the nearest noble gas with atomic number less than the element in question. We subtract the atomic number of that noble gas from the atomic number of the element in question.

The extra electrons we then assign normally starting with using the row after the noble gas ends. We write the name of that noble gas in [brackets] and then write the electronic configuration.

For carbon with Z = 6 the nearest noble gas is Helium. It has the atomic number 2. Subtracting 6 – 2 we get 4 electrons. Helium lies in 1st row. Starting with 2, we get 2s2 2p2.

So the short term electronic configuration is [He] 2s2 2p2

Similarly, for potassium with Z = 19 the nearest noble gas is Argon. It has the atomic number 18. Subtracting 19-18 we get 1 electron. Argon lies in 3rd row. Starting with 4, we get 4s1.

So the short electronic configuration is [Ar] 4s1.

For long term electronic configuration we must write the electronic configuration of the noble gas as well.

So for Carbon it is 1s2 2s2 2p2.

For potassium it is 1s2 2s2 2p6 3s2 3p6 4s1

5 0

3 years ago

These are large chemical compounds, primarily consisting of carbon and hydrogen, that exist in living organisms.

TEA [102]

It is called a Biomolecule

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3 years ago

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Describe the three steps involved in producing crystals of the salt copper sulfate.

denpristay [2]

Answer:

Add copper (II) oxide (insoluble base), a little at a time to the warm dilute sulfuric acid and stir until the copper (II) oxide is in excess (stops disappearing) Filter the mixture into an evaporating basin to remove the excess copper (II) oxide. Leave the filtrate in a warm place to dry and crystallize.

7 0

2 years ago