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natta225 [31]
2 years ago
7

A reaction between an acid and a base produced lithium chloride (LiCl). What acid and base combination could produce this salt?

Chemistry
1 answer:
erica [24]2 years ago
4 0

Answer:

B. hydrochloric acid and lithium hydroxide

Explanation:

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What is a submatomic particle
Amanda [17]

Answer: A subatomic particle is a particle smaller than an atom, hence the name broken down into a sub category

Explanation:

5 0
4 years ago
Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze
Rashid [163]

Answer:

a. ΔH^0_{rxn} = -108.0\frac{kJ}{mol}

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

Compound                    H^0_{t}(kJ/mol)

Liquid EO                       -77.4

CH_4_(g_)                            -74.9                

CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   -108.0 \frac{kJ}{mol}

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}

T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

7 0
4 years ago
A standard soft drink can has a volume of 0.355 L. An empty can is sealed while
beks73 [17]

Answer:

0.35 atm

Explanation:

To solve this problem, we use Boyle's Law: P_1V_1=P_2V_2 , where P is the pressure and V is the volume.

Here, V_1 = 0.355 L, P_1 = 1.0 atm, and V_2 = 0.125 L. So, just plug these values into the equation:

(1.0) * (0.355) = P_2 * (0.125)  ⇒  P_2 ≈ 0.35 atm

Thus, the pressure is 0.35 atm.

Hope this helps!

8 0
3 years ago
Calculate the theoretical yield of ammonia produced by the reaction of 100 g of H2 gas and 200g of N2 gas? 3H2(g) + N2(g)-----&g
Alex787 [66]
Moles of Hydrogen present: 100 / 2 = 50 moles

Moles of Nitrogen present: 200 / 28 = 7.14 moles

Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles

Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.

Molar ratio of Nitrogen : Ammonia = 1 : 2

Moles of ammonia = 7.14 x 2 = 14.28 moles
8 0
4 years ago
Read 2 more answers
Is classified as a substance <br><br> A. Elements and compound <br><br> B.mixtures
Vika [28.1K]
Pretty sure it’s Mixture if I’m not wrong
8 0
3 years ago
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