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Katyanochek1 [597]
3 years ago
8

The arrangement of particles is least ordered in a sample of

Chemistry
1 answer:
lora16 [44]3 years ago
4 0

Answer:

The arrangement of particles is least ordered is  a > c > b > d

Explanation:

Given that,

The arrangement of particles is least ordered in a sample of NaCl.

We know that,

The least ordered arrangement are.

The form of state of particle in least ordered.

So, The least ordered arrangement will be

aqua > gas > liquid > solid

The arrangement of particles is least ordered in a sample of NaCl

NaCl(aq) > NaCl(g) > NaCl(l) > NaCl(s)

Hence, The arrangement of particles is least ordered is a > c > b > d

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What is the base dissociation constant for a weak base at equilibrium, B + H2O <===> BH+ + OH–?
podryga [215]
Depends on what the base is. You would reference the base dissociation chart for that value.
3 0
3 years ago
Read 2 more answers
The average propane cylinder for a residential grill holds approximately 18 kg of propane. How much energy (in kJ) is released b
brilliants [131]
Combustion is a chemical reaction between a fuel and an oxidant, oxygen, to give off combustion products and heat. Complete combustion results when all of the fuel is consumed to form carbon dioxide and water, as in the case of a hydrocarbon fuel. Incomplete combustion results when insufficient oxygen reacts with the fuel, forming soot and carbon monoxide. 

The complete combustion of propane proceeds through the following reaction:
C_{3} H_{8} + 5O_{2} --> 3CO_{2} + 4H_{2}O

Combustion is an exothermic reaction, which means that it gives off heat as the reaction proceeds. For the complete combustion of propane, the heat of combustion is (-)2220 kJ/mole, where the minus sign indicates that the reaction is exothermic. 

The molar mass of propane is 44.1 grams/mole. Using this value, the number of moles propane to be burned can be determined from the mass of propane given. Afterwards, this number of moles is multiplied by the heat of combustion to give the total heat produced from the reaction of the given mass of propane.

 14.50 kg propane  x <u> 1000 g </u> x <u>  1 mole propane   </u>  x <u>  2220 kJ  </u>   
                                     1 kg              44.1 g                     1 mole

= 729,931.97 kJ
8 0
3 years ago
SO2(g) + NO2(g) ↔ SO3(g) + NO(g) Kc = 0.33 A reaction mixture contains 0.41 M SO2, 0.14 M NO2, 0.12 M SO3 and 0.14 M NO. Which o
stepladder [879]

Answer:

The reaction will shift in the direction of products.

Explanation:

<u>Step 1:</u> Data given

A reaction mixture contains:

0.41 M SO2

0.14 M NO2

0.12 M SO3

0.14 M NO

<u>Step 2:</u> The balanced equation

O2(g) + NO2(g) ↔ SO3(g) + NO(g)      Kc = 0.33

<u>Step 3:</u> Define the direction of the shift of  reaction:

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants.

When Q=K, the system is at equilibrium and there is no shift to either the left or the right.

<u>Step 4:</u> Calculate Q

Q = [NO][SO3]/[SO2][NO2]

Q = (0.14 *0.12)/(0.41*0.14)

Q = 0.0168/0.0574

Q = 0.293

Q<Kc

This means there are more reactants than products. Thud, some of the reactants will become products, causing the reaction to shift to the right.

The reaction will shift in the direction of products.

8 0
3 years ago
A 6.165 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 10.27 grams of CO2 and 3
e-lub [12.9K]

Answer:

Explanation:

mass of carbon in 10.27 g of CO₂ = 12 x 10.27 / 44 = 2.80 g

mass of hydrogen ( H ) in 3.363 g of H₂O = 2  x 3.363 / 18

= .373 g

These masses would have come from the sample of 6.165 g .

Rest of 6.165 g of sample is oxygen .

So oxygen in the sample = 6.165 - ( 2.8 + .373 ) = 2.992 g

Ratio of C  , H , O in the sample

2.8 : .373 : 2.992

C: H : O : : 2.8 : .373 : 2.992

Ratio of moles

C: H : O : : 2.8/12 : .373/1 : 2.992 / 16

C: H : O : : .2333 : .373 : .187

C: H : O : : .2333/.187 : .373/.187 : .187/.187

C: H : O : : 1.247 : 1.99 : 1

C: H : O : : 5 : 8 : 4 ( after multiplying by 4 )

Hence empirical formula

C₅H₈O₄

Molecular formula ( C₅H₈O₄ )n

n ( 5 x 12 + 8 x 1 + 4 x 16 ) = 132

n x ( 60 + 8 + 64 ) = 132

n = 1

Molecular formula = C₅H₈O₄.

3 0
3 years ago
A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
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