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3 years ago

Use the chart to determine which pairs of atoms are arranged in the correct order according to increasing bond polarity?

Chemistry

2 answers:

Mrrafil [7]3 years ago

7 0

Polo of chemical reactions

evablogger [386]3 years ago

6 0

Atoms bond with other atoms to form bonds. in order to do that, they release and electron

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Can someone help me out with this please

Dmitriy789 [7]

Answer:

$molar \: mass \: of \: copper(ii)nitrate = 64 + (14 + 48) \times 3 \\ = 250 \: g \\ 64 \: g \: of \: copper \: produces \: 250 \: g \: of \: copper \: nitrate \\ 10.36 \: g \: of \: copper \: will \: produce \: ( \frac{10.36 \times 250}{64} ) \: g \\ = 40.7 \: g$

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3 years ago

Which words in the sentence are the adjective clause? Alessandra, whom everyone calls "Al," is the best tennis player on the tea

Alinara [238K]

The answer is D. because that makes more sense

4 0

3 years ago

What is the mass of 3.8 × 1024 atoms of argon (Ar)?

gulaghasi [49]

40 g Ar.......6.023*10^23 atoms
x g Ar.......3.8*10^24 atoms
x=40*3.8*10^24/(6.023*10^23)=252,365g Ar

6 0

3 years ago

a 1.25g sample of ore containing iron pyrite (FeS2) was pulverized and ignited in air, converting the FeS2 to Fe2O3 and SO2(g).

svp [43]

Answer:

28.9%

Explanation:

Let's consider the following balanced equation.

2 FeS₂ + 11/2 O₂ ⇒ Fe₂O₃ + 4 SO₂

We can establish the following relations:

The molar mass of Fe₂O₃ is 159.6 g/mol
1 mole of Fe₂O₃ is produced per 2 moles of FeS₂
1 mole of Fe is in 1 mole of FeS₂
The molar mass of Fe is 55.84 g/mol

The amount of Fe in the sample that produced 0.516 g of Fe₂O₃ is:

$0.516gFe_{2}O_{3}.\frac{1molFe_{2}O_{3}}{159.6gFe_{2}O_{3}} .\frac{2molFeS_{2}}{1molFe_{2}O_{3}} .\frac{1molFe}{1molFeS_{2}} .\frac{55.84gFe}{1molFe} =0.361gFe$

The percent of Fe in 1.25 g of the ore is:

$\frac{0.361g}{1.25g} .100\%=28.9\%$

4 0

3 years ago

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

Answer: The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$ .......(1)

Calculating for first dilution:

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

Calculating for second dilution:

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

3 years ago