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BartSMP [9]
3 years ago
6

How many moles of Al2(so4)3 can be produced from 7.25E24

Chemistry
1 answer:
rodikova [14]3 years ago
5 0

12.04moles

Explanation:

Given parameters:

Number of molecules = 7.25 x 10²⁴molecules

unknown:

Number of moles = ?

Solution:

The moles of a substance is the amount of substance that contains avogadro's number of particles:

  1 mole of a substance = 6.02 x 10²³

   x moles of the compound will contain: \frac{7.25 x 10[tex]^{24}}{6.02 x 10^{23}}[/tex] = 12.04moles

 This gives 12.04moles

Learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

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Is oxygen a beginning substance or ending substance?
stiks02 [169]

Answer:

Explanation:

The substances that are present at the beginning are called reactants and the substances present at the end are called products. Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products. Reactants are the starting materials, that is, whatever we have as our initial ingredients.Oxygen, a colorless, odorless, tasteless gas essential to living organisms, being taken up by animals, which convert it to carbon dioxide; plants, in turn, utilize carbon dioxide as a source of carbon and return the oxygen to the atmosphere. Oxygen forms compounds by reaction with practically any other element.

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3 years ago
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In the reaction H2CO3 (aq) + 3NH3 (aq) = 2 NH4+ (aq) + CO3 2-, how many electrons are transferred?
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3 years ago
Unknown element has two isotopes. Isotope A has a mass of 34 amu and abundance of 52%, isotope B has a mass of 33 amu and abunda
ZanzabumX [31]

Answer:

x = 33.52 amu

Explanation:

It is given that,

Isotope A has a mass of 34 amu and an abundance of 52%, isotope B has a mass of 33 amu and an abundance of 48%.

Let x is the average atomic mass of this element. It can be calculated as follows :

x=52\%\ \text{of}\ 34+48\%\ \text{of}\ 33\\\\x=\dfrac{52}{100}\times 34+\dfrac{48}{100}\times 33\\\\x=0.52\times 34+0.48\times 33\\\\x=33.52\ \text{amu}

So, the average atomic mass of this element is 33.52 amu.

4 0
3 years ago
A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.
seropon [69]

<u>Answer:</u> The p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

<u>Explanation:</u>

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

  • Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, M_1=(2\times 0.0031)=0.0062M

  • Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, M_2=(2\times 0.0042)=0.0084M

To calculate the concentration of nitrate ions in the solution, we use the equation:

M=\frac{M_1V_1+M_2V_2}{V_1+V_2}

Putting values in above equation, we get:

M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M

Calculating the p-function of zinc ions and nitrate ions in the solution:

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\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]

\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51

  • <u>For nitrate ions:</u>

\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]

\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14

Hence, the p-function of Zn^{2+} and NO_3^{-} ions are 2.51 and 2.14 respectively.

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What does conserving mass mean in a chemical equation?
Keith_Richards [23]
The answer would be C. Hope this helps
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