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3 years ago

The graph given below shows the motion of four runners P, Q, R, and S in a 5 km marathon.

Chemistry

1 answer:

wel3 years ago

3 0

Answer:-

As we can see from the graphical data,

The distance covered by all the four runners is the same 5 km.

Among the four athletes, Athlete P covers the distance in under three hours.

It is the minimum time taken among the four athletes.

Thus Athlete P covers the 5 km distance in the minimum amount of time.

We know that speed = $\frac{Distance}{Time}$

Since time taken for P is minimum, his speed is the maximum. P ran the fastest.

Time taken by Q = 4.5 hours.

Speed of Q = $\frac{Distance}{Time taken by Q}$

= $\frac{5km}{4.5hour}$

= 1.1 km/ hr

Time taken by R = 6 hours.

Speed of R = $\frac{Distance}{Time taken by R}$

= $\frac{5km}{6hour}$

= 0.8 km/ hr

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Jill then wrote the formulas for all the possible compounds that sodium can form with the other ions. What formulas did she writ

sp2606 [1]

Explanation:

Sodium has atomic number of 11 and its electronic configuration is given by:

$[Na]=1s^22s^22p^63s^1$

The nearest stable electronic configuration to sodium is of the neon. So, in order to attain stability of noble gas it will loose its single electron.

$Na\rightarrow Na^++1e^-$

$[Na^+]=1s^22s^22p^63s^0$

Sodium has single valency that is 1.

Let nbe the valency of the ion 'X'

By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

$Na_nX_1$

So, the formulas for all the possible compounds that sodium can form with the other ions will be:

$Na_nX_1$

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3 years ago

22-How many molecules are there in 4L of oxygen at 500C° and 0.0658 atm

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Answer:

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7 0

3 years ago

Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L

Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___Cu+² __|_2OH-____|

|Initial concentration(M)|___0__|_0______|

|Change in concentration(M)|_+S |__+2S__|

|Equilibrium concentration(M)|_S _|2S___|

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

$s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}$

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

$Cu (OH)2 = \frac{1.8 \times {10}^{ - 7} mol \:Cu (OH)2 }{1L} \times \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2} \\ = 1.75428 \times 10 ^{ - 5}$

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

8 0

3 years ago

Most gases exist as molecules, for example, fluorine is F2.

Kryger [21]

They are all stable and have eight valence electrons

4 0

3 years ago

Read 2 more answers

What volume of O2 collected at 22.0 and 728 mmHg would be produce by the decomposition of 8.15 g KClO3?

adell [148]

Answer:

There is 2.52 L of O2 collected

Explanation:

Step 1: Data given:

Temperature = 22.0 °C

Pressure = 728 mmHg = 728 /760 = 0.958 atm

Mass of KClO3 = 8.15 grams

Molar mass of KClO3 = 122.55 g/mol

Step 2: The balanced equation

2KClO3(s) → 2KCl(s) + 3O2(g)

Step 3: Calculate moles of KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3= 8.15 grams / 122.55 g/mol

Moles KClO3 = 0.0665 moles

Step 4: Calculate moles of O2

For 2 moles of KClO3 we'll have 2 moles of KCl and 3 moles of O2 produced

For 0.0665 moles of KClO3 we have 3/2 * 0.0665 = 0.09975 moles

Step 5: Calculate vlume of O2

p*V = n*R*T

V = (n*R*T)/p

⇒ with n = the number of moles O2 = 0.09975 moles

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = 22.0 °C = 273 +22 = 295 Kelvin

⇒ with p = 0.958 atm

V = (0.09975 * 0.08206 * 295) / 0.958

V = 2.52 L

There is 2.52 L of O2 collected

7 0

3 years ago