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3 years ago

Please answer each:) <3

1 answer:

Nutka1998 [239]3 years ago

4 0

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The diameter of a hydrogen atom is 106 pm. Find the length in kilometers of a row of 4.34 x 1023 hydrogen atoms?

valina [46]

The length in kilometers of a row of 4.34 x 1023 hydrogen atoms is 4.34 x 10¹⁴ km.

<h3>Length of the entire hydrogen atoms</h3>

The length of the entire hydrogen atom is calculated as follows;

Length of the row = number x diameter of one

Length of the row = 4.34 x 10²³ x 10⁶ x 10⁻¹²

Length of the row = 4.34 x 10¹⁷ m

Length of the row = 4.34 x 10¹⁴ km

Thus, the length in kilometers of a row of 4.34 x 1023 hydrogen atoms is 4.34 x 10¹⁴ km.

Learn more about diameter of hydrogen atom here: brainly.com/question/13796082

#SPJ1

3 0

2 years ago

Please explain how to do it as well!

kogti [31]

Answer:

a. $\bold{C_{6}H_{12}O_{6}\rightarrow 6CO_{2}+6H_{2}O}$

△H=−72 kcal

The energy required for production of 1.6 g of glucose is [molecular mass of glucose is 180 gm]

$\bold{Fe_{2}(SO_{4})_{3}+3Ba(OH)_{2}\rightarrow 3BaSO_{4}+2Fe(OH)_{3}}$

The iron(III) ions and chloride ions remain aqueous and are spectator ions in a reaction that produces solid barium sulfate.

7 0

3 years ago

Read 2 more answers

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

The natural abundances of elements in the human body, expressed as percent by mass, are oxygen (o), 65 percent; carbon (c), 18 p

Vlada [557]

It is given that the person weighs 62 kg = 62,000 g

Natural abundances in mass percent are:

O = 65%

C = 18%

H = 10%

N = 3.0%

Ca = 1.6%

P = 1.2%

Corresponding weights of the elements are:

O = 65/100 * 62000 g = 40.30 * 10^3 g

C = 18/100 * 62000 g = 11.16 * 10^3 g

H = 10/100 * 62000 g = 62.00 * 10^2 g

N = 3.0/100 * 62000 g = 18.60 * 10^2 g

Ca = 1.6/100 * 62000 g = 9.92 * 10^2 g

P = 1.2/100 * 62000 g = 7.44 * 10^2 g

3 0

3 years ago

Read 2 more answers

6. According to Crystal Field Theory, what should be the color of an aqueous solution of the ion [Y(H2O)6]3

UkoKoshka [18]

Answer:

Colourless

Explanation:

We know that Y^3+ has the electronic configuration of;

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 (the 5s and 4d levels are empty).

According to the crystal field theory, the colour of complexes result from transitions between incompletely filled d orbitals.

As a result of this, complexes with empty or completely filled d orbitals are colourless. Thus, [Y(H2O)6]3 is colourless according to the Crystal Field Theory.

4 0

3 years ago