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anygoal [31]
2 years ago
15

For which of these processes is the value of ΔH expected to be negative?I. The temperature (of water) increases when calcium chl

oride dissolves in waterII. Steam condenses to liquid waterIII. Water freezesIV. Dry ice sublimesA.) IV onlyB.) I, II, and IIIC.) I onlyD.) II and III only
Chemistry
1 answer:
Marina CMI [18]2 years ago
6 0

Answer: B.) I, II, and III

Explanation:

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and \Delta H for the reaction comes out to be negative.

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and \Delta H for the reaction comes out to be positive.

I) The temperature (of water) increases when calcium chloride dissolves in water : Thus the reaction is exothermic and  \Delta H for the reaction comes out to be negative.

II)  Steam condenses to liquid water : The energy is released when bonds are formed when it coverts from gas to liquid and thus \Delta H for the reaction comes out to be negative.

III) Water freezes : The energy is released when bonds are formed to get converted from liquid to solid and thus \Delta H for the reaction comes out to be negative.

IV) Dry ice sublimes : The energy is absorbed when bonds are broken to get converted from solid to gas and thus \Delta H for the reaction comes out to be positive.

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What volume of water should be used to dissolve 19.6 g of LiF to create a 0.320 M solution?
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Answer:

2.4 litters of water are required.

Explanation:

Given data:

Mass of LiF = 19.6 g

Molarity of solution = 0.320 M

Volume of water used = ?

Solution:

Number of moles = mass/molar mass

Number of moles = 19.6 g/ 26 g/mol

Number of moles = 0.75 mol

Volume required:

Molarity = number of moles/ volume in L

Now we will put the values in above given formula.

0.320 M = 0.75 mol /  volume in L

Volume in L = 0.75 mol  /0.320 M

     M = mol/L

Volume in L = 2.4 L

4 0
2 years ago
Industrial production of nitric acid, which is used in many products including fertilizers and explosives, approaches 10 billion
mylen [45]

Answer: 9.361\times 10^{4} kJ

Explanation:

The balanced chemical equation :

4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)  \Delta H^0_{rxn}=-902.0kJ

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.056\times 10^3g}{17g/mol}=415.1moles

According to stoichiometry:

4 moles of NH_3 produces = 902.0 kJ of energy

415.1 moles of NH_3 produces =\frac{902.0}{4}\times 415.1=9.361\times 10^{4} kJ of energy

Thus the change in enthalpy is 9.361\times 10^{4} kJ

5 0
2 years ago
According to periodic properties, what would be the most likely formula for the product obtained when Lv reacts with H2(g)?
svetoff [14.1K]

Answer:

H₂Lv

Explanation:

Lv is at group 6 on the periodic table, so it has 6 valence electrons, likely oxygen. Thus, to be stable, it needs to gain 2 electrons. Hydrogen has 1 electron in its valence shell, so H₂ can share 2 electrons with Lv, and because of that, the product would be:

H₂Lv.

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2 years ago
In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o
Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

The specific gravity of the wood chips is 0.494.

The average volume of a log is

V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2}  \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}

The weight of one log is

M=\rho*V=0.494*\rho_w*12.57  ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm

To provide 3617 ton/day of wood chips, we need

n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

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