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expeople1 [14]
3 years ago
9

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to rea

ct 37.1 g of potassium chlorate () with excess red phosphorus, what mass of tetraphosphorus decaoxide () could be produced
Chemistry
1 answer:
Alexus [3.1K]3 years ago
8 0

25.55 grams of tetraphosphorus decaoxide could be produced by the reaction.

Explanation:

First the balanced chemical reaction of the production of tetraphosphorus decaoxide is to be known.

The chemical equation is

10 KClO3 + 3P4⇒ 3 P4010 + 10 KCl

The number of moles of KCLO3 will be calculated by the formula:

number of moles = mass of the compound given ÷ atomic mass of the compound

n = 37.1 ÷ 122.55     ( atomic mass of KClO3 is 122.55 gm/mole)

   = 0.30 moles

From the stoichiometry

10 moles of KClO3 is required to produce 3 moles of P4O10

when 0.30 moles of KClO3 is used x moles of P4O10 is formed

thus, 3 ÷ 10 = x ÷ 0.30

   = 0.09 moles of KClO3 is produced

To know the mass of P4O10 apply the formula

mass = number of moles × atomic mass

        = 0.09 × 283.886   ( atomic mass of P4O10 is 283.88 gram/mole)

         = 25.55 grams of P4O10 could be produced.

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Give the result of the following mathematical operations to the correct number of significant figures. 5.659 * (3.496 - 2.814) =
shtirl [24]

Answer:

The result is 3.859 in which we use four significant figures.

Explanation:

We start by solving the mathematical operation :

5.659.(3.496-2.814)=5.659(0.682)=3.859438

The result for the operation is 3.859438 but the numbers in the operation are given with four significant figures and that is why we are going to use four significant figures to express the result

To express 3.859438 with four significant figures we use the first four digits that appear from left to right starting by the first digit that is different to zero

In this case : 3.859 will be the result with four significant figures.

We also use a rule that says : To decide if the last significant figure remains the same we look for the value of the digit at its right.

If that number is greater than or equal to 5 ⇒ we sum one to the last significant figure

For example 3.859738 = 3.860 with four significant figures because the ''7'' is greater that 5

If that number is less than 5 ⇒ the last significant figure remains the same

In our case : 3.859438 = 3.859 because ''4'' is less than ''5''

7 0
3 years ago
Explain how sand is form​
Reptile [31]

Answer:

Sand comes from many locations, sources, and environments. Sand forms when rocks break down from weathering and eroding over thousands and even millions of years. Rocks take time to decompose, especially quartz (silica) and feldspar .

Explanation:

7 0
3 years ago
Wherever there is an action force, there must be a reaction force that
Fittoniya [83]

Answer:

is equal to the action force , but acts in the opposite direction.

Explanation:

  • All you need to solve is apply Newton's third law.

It states that

  • Every action has an equal and opposite reaction

Or

\\ \tt\bull\rightarrowtail F_A=-F_A

4 0
2 years ago
Based on formal charges, draw the most preferred Lewis structure for the chlorate ion, ClO3−. To add lone pairs, click the butto
vova2212 [387]

Answer:

The structure is shown in the diagram.

Explanation:

Lewis structure : In order to draw Lewis structure we will calculate the total number of valence electrons in the molecule.

The valence electrons from Cl : 7

The valence electrons from O = 3 X 6 = 18

Charge negative so more electrons = 2

total electrons = 7 + 18 +2 = 27

Now we will distribute the electrons on each atom and in between atoms as shown in the diagram.

8 0
3 years ago
Read 2 more answers
Carbon-14 has a half-life of 5,700 years.
Maksim231197 [3]

Answer:

3.106\ \text{g}

Explanation:

t_{1/2} = Half-life of carbon = 5700 years

t = Time at which the remaining mass is to be found = 10400 years

m_0 = Initial mass of carbon = 11 g

Decay constant is given by

\lambda=\dfrac{\ln2}{t_{1/2}}

Amount of mass remaining is given by

m=m_0e^{-\lambda t}\\\Rightarrow m=m_0e^{-\dfrac{\ln2}{t_{1/2}} t}\\\Rightarrow m=11e^{-\dfrac{\ln 2}{5700}\times 10400}\\\Rightarrow m=3.106\ \text{g}

The amount of the substance that remains after 10400 years is 3.106\ \text{g}.

5 0
3 years ago
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