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3 years ago

Hydroelectricity and tidal energy are similar because

Chemistry

1 answer:

vovikov84 [41]3 years ago

3 0

C. both sources of energy rely on water.

Explanation:

Hydroelectricity and tidal energy are similar energy sources because they both rely on water.

They harness the power of moving water to produce their energy.

Tidal energy is the energy derived from moving water as crashing waves in an ocean.
Hydroelectricity is produce when water from a dam channel is used to drive heavy turbines.
They are both renewable energy sources.

learn more:

Non-renewable resources brainly.com/question/2948717

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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

Is MgCO3 organic or inorganic

schepotkina [342]

Its inorganic as MgCO3 is contains no carbon more hydrogen which is a crutial component of all organic compounds

4 0

3 years ago

Consider the redox reaction below.

vovangra [49]

Answer:

Zn(s) → Zn⁺²(aq) + 2e⁻

Explanation:

Let us consider the complete redox reaction:

Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

This is a redox reaction because, both oxidation and reduction is simultaneously taking place.

Oxidation (loss of electrons or increase in the oxidation state of entity)
Reduction (gain of electrons or decrease in the oxidation state of the entity)

An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet configuration. An octet configuration is that of outer shell configuration of noble gas.

Here Zn(s) is undergoing oxidation from OS 0 to +2

And H in HCl (aq) is undergoing reduction from OS +1 to 0.

Therefore, for this reaction;

Oxidation Half equation is:

Zn(s) → Zn⁺²(aq) + 2e⁻

Reduction Half equation is:

2H⁺ + 2e⁻ → H₂(g)

4 0

3 years ago

What is the atom inventory for the following equation after it is properly balanced? ____NaOH + ____CuCl2 Imported Asset ____NaC

faltersainse [42]

Answer:

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2;

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

Explanation:

NaOH + CuCL2 —> NaCl + Cu(OH)2

The balanced equation can be achieved by doing the following:

There are 2 oxygen and 2 hydrogen atom on the right side. This is balanced by putting 2 in front of NaOH as shown below:

2NaOH + CuCL2 —> NaCl + Cu(OH)2

This makes Na to be unbalanced. Now to balance Na, put 2 in front of NaCl as illustrated below

2NaOH + CuCL2 —> 2NaCl + Cu(OH)2

Now the equation is balanced.

Reactants: Na = 2, O = 2, H = 2, Cu = 1, Cl = 2

Products: Na = 2, Cl = 2, Cu = 1, O = 2, H = 2

4 0

3 years ago

Calculate the ph of a 0.005 m solution of potassium oxide k2o

Alecsey [184]

First, we have to see how K2O behaves when it is dissolved in water:

K2O + H20 = 2 KOH

According to reaction K2O has base properties, so it forms a hydroxide in water.
For the reaction next relation follows:

c(KOH) : c(K2O) = 1 : 2

So,

c(KOH)= 2 x c(K2O)= 2 x 0.005 = 0.01 M = c(OH⁻)

Now we can calculate pH:

pOH= -log c(OH⁻) = -log 0.01 = 2

pH= 14-2 = 12

3 0

3 years ago