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3 years ago

Word equation for Cu(s)+O2(g)=CuO(s)

Chemistry

2 answers:

Mazyrski [523]3 years ago

7 0

Answer:

Copper solid plus oxygen gas giving solid cupric oxide

Explanation:

The given reaction is:

$Cu(s) +O2(g)\rightarrow CuO(s)$

The balanced equation is:

$2Cu(s) +O2(g)\rightarrow 2CuO(s)$

The formula name of the reactants and products are:

Cu (s) = copper in the solid phase

O2 (g) = oxygen in the gas phase

CuO (s) = copper (II) oxide or cupric oxide

The word equation would be:

Copper solid plus oxygen gas giving solid cupric oxide

Ludmilka [50]3 years ago

4 0

Copper + Oxygen = Copper Oxide

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PSYCHO15rus [73]

Answer:

iconic compound is the answer

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3 years ago

Consider the reaction 2CO(g) + O2(g)2CO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

Marizza181 [45]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})]-[(1\times \Delta S^o_{(O_2(g))})+(2\times \Delta S^o_{(CO(g))})]$

We are given:

$\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(CO)}=197.674J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (213.74))]-[(1\times (205.14))+(2\times (197.674))]\\\\\Delta S^o_{rxn}=-173.008J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-173.008) J/K = 173.008 J/K

We are given:

Moles of CO gas reacted = 2.25 moles

By Stoichiometry of the reaction:

When 2 moles of CO is reacted, the entropy change of the surrounding will be 173.008 J/K

So, when 2.25 moles of CO is reacted, the entropy change of the surrounding will be = $\frac{173.008}{1}\times 2.25=432.52J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of CO is reacted is 432.52 J/K

3 0

3 years ago

Why is the cellulose part of human diet

Luden [163]

Answer:

Cellulose stimulates the work of intestines and increases their discharge.

Explanation:

Cellulose is a polysaccharide made of glucose polymers, which build plant fibers. Because of this, the plants possess their firmness.

Humans do not digest cellulose because they do not possess certain enzymes, but it is useful for their digestion because it stimulates the work of the intestines.

Cellulose enhances intestinal peristalsis. Peristalsis is series of contractions that allow the progressive movement of food through the digestive tract.

4 0

2 years ago

A hot air balloon starts with its temperature at 68.7°C and a pressure of 0.987 ATM and volume of 564L at what temperature in de

ICE Princess25 [194]

Answer:

54.7°C is the new temperature

Explanation:

We combine the Ideal Gases Law equation to solve this.

P . V = n. R. T

As moles the balloon does not change and R is a constant, we can think this relation between the two situations:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

T° is absolute temperature (T°C + 273)

68.7°C + 273 = 341.7K

(0.987 atm . 564L) / 341.7K = (0.852 atm . 625L) / T₂

1.63 atm.L/K = 532.5 atm.L / T₂

T₂ = 532.5 atm.L / 1.63 K/atm.L → 326.7K

T° in C = T°K - 273 → 326.7K + 273 = 54.7°C

3 0

3 years ago

Which of the following is a galvanic cell?

sladkih [1.3K]

C. Aluminum (Al) oxidized, zinc (Zn) reduced

<h3>Further explanation</h3>

Given

Metals that undergo oxidation and reduction

Required

A galvanic cell

Solution

The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.

$\large {\boxed {\bold {E ^ ocell = E ^ ocatode -E ^ oanode}}}$

or:

E ° cell = E ° reduction-E ° oxidation

For the reaction to occur spontaneously (so that it E cell is positive), the E° anode must be less than the E°cathode

If we look at the voltaic series:

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au

The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)

From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.

7 0

3 years ago

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