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3 years ago

How many valence electrons are in an atom of gold?

1 answer:

3 0

Chemical reactivity of all of the different elements in the periodic table depends on the number of electrons in that last, outermost level, called the valence level or valence shell. In the case of gold, there is only one valence electron in its valence level. Hope I helped! :)

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A chemist prepares a solution of sodium chloride(NaCl) by measuring out 25.4g of sodium chloride into a 100ml volumetric flask a

labwork [276]

Answer: 4 molL-1

Explanation:

Detailed solution is shown in the image attached. The number of moles of NaCl is first obtained. Since the molarity must be in units of molL-1, the volume is divided by 1000 and the formula stated in the solution is applied and the answer is given to one significant figure.

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3 years ago

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The lab procedure mentions that not all of the oxone may dissolve after being added to the water because the water may be satura

ollegr [7]

Answer:

The answer is (C) There are more solute molecules than water molecules.

Explanation:

A saturated solution is one in which no more solute can be dissolved or disintegrated into the solvent. When or if the ozone stops being dissolved in the water, it implies that the water has already taken on more ozone molecules than it can contain, meaning there are more solute molecules (ozone molecules) than there are solvent molecules (water molecules).

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3 years ago

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Question 1 Which of the following is a false

Bogdan [553]

Answer:

1.A 2.C 3.B

4. d

Explanation:

6 0

3 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

Combustibilty defined?

Phantasy [73]

Capable of catching fire

8 0

3 years ago