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1 year ago

The molar mass of silicon (Si) is 28.09 g/mol. Calculate the number of atoms in a 92.8 mg sample of Si.

Chemistry

1 answer:

adelina 88 [10]1 year ago

6 0

Explanation:

one mole of any substance there are 6.022×1023 units of that substance. (This number is called Avogadro's number, NA.)

We need to convert the mass of silicon to moles using the molar mass of silicon, 28.06gmol. This number means that one mole of pure silicon would have a mass of 28.06g. Our given mass, however, is in milligrams; to convert this to grams we'll use the conversion factor 1g103mg:

5.86mg Si(1g103mg)=0.00586g Si

Now, using silicon's molar mass, we'll convert this mass to moles of Si:

0.00586g Si(1mol Si28.06g Si)=2.09×10−4mol Si

Finally, let's use Avogadro's number to convert

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Lead is malleable, so it can be pounded into flat sheets without breaking. How does the bonding within lead help to explain this

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The answer is Metallic bonds involve many valence electrons shared by many atoms, so the bonds can move around as the metal is pounded. The metallic bond structure of lead forms a cubic crystal structure and the atoms can roll over one another without breaking the metallic bonds. This is especially because the p orbital electrons of lead can be delocalized and the electrons can be shared with other lead ions in the cubic structure of lead.

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2 years ago

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How many moles are in 19.6 g of Sodium (Na)? And the conversion factor?

Snezhnost [94]

Answer:

n=0.852 moles

Explanation:

Given mass is, m = 19.6 g

The molar mass of sodium is, M = 22.99 u

We need to find the no of moles in 19.6 of Sodium. We know that, no of moles is equal to given mass divided by molar mass.

$n=\dfrac{m}{M}\\\\n=\dfrac{19.6}{22.99}\\\\n=0.852$

So, there are 0.852 moles in 19.6 g of Sodium.

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2 years ago

Tumu’s class was given an assignment to feature a scientist that contributed to the development of the cell theory. The class de

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Answer: the way root cells reproduce to increase root length

Explanation:

Rudolf Ludwig Carl Virchowas known as the founder of social medicine and also the father of modern pathology.

Virchow posited that all cells are gotten from already existing cells and he used this in his work towards cellular pathology, as it was made clear that diseases takes place at the cellular level. He posited that the cells that are malfunctioning cause diseases.

Based on the above analysis, the image that Tumu would most likely use in his assignment to feature Rudolf Virchow is the way root cells reproduce to increase root length.

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1 year ago

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Calcium reacts with bromine to form calcium bromide. 2.1 Which atom will lose electrons? 2.2 How many electrons will it lose? 2.

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Answer:

For 2.1: Calcium atom will loose electrons.

For 2.2: It will loose 2 electrons.

For 2.3: Bromine atom will gain electrons.

For 2.4: It will gain 1 electron.

For 2.5: The ions formed are $Ca^{2+}\text{ and }Br^-$

For 2.6: The chemical formula is $CaBr_2$

Explanation:

Calcium bromide is formed by the combination of calcium and bromine atoms. This compound is ionic compound because electrons get transferred from one atom to another atom.

Calcium is the 20th element of the periodic table. It is a metal and it looses its electrons. The electronic configuration for this element is $1s^22s^22p^63s^23p^64s^2$

This element will loose 2 electrons and forms $Ca^{2+}$ ion.

Bromine is the 35th element of the periodic table. It is a non-metal and it gains electrons. The electronic configuration for this element is $[Ar]4s^23d^{10}4p^5$

This element will gain 1 electron and forms $Br^-$ ion.

The chemical formula for the given compound is $CaBr_2$

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1 year ago

At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.400 M . N 2 ( g )

stiks02 [169]

Answer: The equilibrium concentration of NO after it is re-established is 0.55 M

Explanation:

For the given chemical equation:

$N_2(g)+O_(g)\rightleftharpoons 2NO(g)$

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=\frac{[NO]^2}{[N_2][O_2]}$ .....(1)

We are given:

$[NO]_{eq}=0.400M$

$[N_2]_{eq}=0.200M$

$[O_2]_{eq}=0.200M$

Putting values in expression 1, we get:

$K_{eq}=\frac{(0.400)^2}{0.200\times 0.200}\\\\K_{eq}=4$

Now, the concentration of NO is added and is made to 0.700 M

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

The equilibrium will shift in backward direction.

$N_2(g)+O_(g)\rightleftharpoons 2NO(g)$

Initial: 0.200 0.200 0.700

At eqllm: 0.200+x 0.200+x 0.700-2x

Putting values in expression 1, we get:

$4=\frac{(0.700-2x)^2}{(0.200+x)\times (0.200+x)}\\\\x=0.075$

So, equilibrium concentration of NO after it is re-established = (0.700 - 2x) = [0.700 - 2(0.075)] = 0.55 M

Hence, the equilibrium concentration of NO after it is re-established is 0.55 M

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