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3 years ago

How does water's dissolving power support life on Earth?

1 answer:

6 0

The dissolving power of water is very important for life on Earth. Wherever water goes, it carries dissolved chemicals, minerals, and nutrients that are used to support living things. Because of their polarity, water molecules are strongly attracted to one another, which gives water a high surface tension

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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:

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

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

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

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

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (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

If a buffer solution is 0.190 m in a weak acid (ka = 8.2 × 10-5) and 0.590 m in its conjugate base, what is the ph?\

Lubov Fominskaja [6]

You use the Henderson - Hasselbalch equation pH = pKa + log ([salt]/[acid]) pKa = -log (8.2*10^-5) = 4.081 pH = 4.081 + (0.590/0.190) pH = 4.081 + log 3.105 pH = 4.081 + 0.49206 pH = 4.573

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3 years ago

Aleks [24]

Electron microscopes differ from light microscopes in that they produce an image of a specimen by using a beam of electrons rather than a beam of light. Electrons have much a shorter wavelength than visible light, and this allows electron microscopes to produce higher-resolution images than standard light microscopes

3 0

3 years ago

When a substance undergoes a chemical change, the product exhibits different physical properties than the starting material. Des

elena55 [62]

Answer:

I) Change in solubility

II) Change in boiling point

III) Change in colour

Explanation:

A chemical change involves formation of new products and is not reversible.

So, once two liquid solutions are mixed and a chemical change takes place, the new product will have the following:

- a new solubility rate, i.e it will dissolve at a rate different from the two liquid solution

- a new boiling point i.e it takes a new point at which its molecules liberate to yield vapour

- a new colour might be detected, as the individual solution each has its own colour

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3 years ago

Can someone answer 4 for me please?

Mice21 [21]

I can’t see the whole question......

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3 years ago