Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = + 384.3 k J / m o l what is Δ H for this reaction? X ( s ) + 1 2 O 2 ( g ) + CO 2 ( g ) ⟶ XCO 3 ( s )

Answers

Answer 1

Answer:

Answer: The $\Delta H^o_(rxn)$ for the reaction is -1052.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+(1)/(2)O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_(rxn)=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+(1)/(2)O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_(rxn)=[1* \Delta H_1]+[1* (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_(rxn)=[(1* (-668.5))+(1* (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_(rxn)$ for the reaction is -1052.8 kJ.

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What are two functions of the cilia?

Answers

Answer:

- Proper urine flow by signalling the kidney cells.

- They act as mechanoreceptors or sensory receptors.

Explanation:

Motile' (or moving) cilia are found in the lungs, respiratory tract and middle ear. These cilia have a rhythmic waving or beating motion. They work, for instance, to keep the airways clear of mucus and dirt, allowing us to breathe easily and without irritation. They also help propel sperm.
Btw I don’t know if this is what you meant

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74 atm. When the experiment is repeated using 0.020 mol of ethanoic acid instead of methyl methanoate, the measured pressure is lower than 0.74 atm. The lower pressure for ethanoic acid is due to the following reversible reaction.CH3COOH(g)+CH3COOH(g) ⇋ (CH3COOH)2(g)+Assume that when equilibrium has been reached, 50 percent of the ethanoic acid molecules have reacted.i. Calculate the total pressure in the vessel at equilibrium at 450 K.ii. Calculate the value of the equilibrium constant, Kp, for the reaction at 450 K

Answers

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

= 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product $(CH3COOH)^(2)$ gas formed are calculated as

0.010 mol CH3COOH * 1 mol $(CH3COOH)^(2)$ / 2 mol CH3COOH

= 0.005 mol $(CH3COOH)^(2)$

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^(2)$

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

= 0.74 atm * 0.015 mol / 0.020 mol

= 0.555 atm

16. Which of the following properties of the elements remain unchanged across#period
A lonivation energy
C Electro negativity
13. Number of shells
D. Nuclear charge

Answers

Answer:

Number of shells

Explanation:

Across the period on the periodic table, the number of shells remains the same. There is no variation or trend for the number of shells on a period.

The number of shell is the energy level corresponding to an atom.
As we go from period to period, the energy level increases by 1.
The period number corresponds to the number of shells.
Elements in period 4 will have 4 shells.
There is no variation of any sort in the number of shell across a period.

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine ________.

Answers

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

Answer;

-past temperatures

The ratio of oxygen-16 and oxygen-18 isotopes in plankton fossils in deep-sea sediments can be used to determine past temperatures.

Explanation;

-O-16 will evaporate more readily than O-18 since it is lighter, therefore; during a warm period, the relative amount of O-18 will increase in the ocean waters since more of the O-16 is evaporating.

-Hence, looking at the ratio of O16 to O18 in the past can give clues about global temperatures.

. Explain why, in the sample calculations, 0.1 g of the unknown produced a GREATER freezing point depression than~e same mass of naphthalene.