Compare the arrangements of individual particles in solids, liquids, and gases?

1 answer:

4 0

Solid particles are all compacted together, liquid particles are not as close as solid particles but still move together, and gas particles do not stay together

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A student mixes some sugar into a cup of hot water. The sugar dissolves in the liquid. What are the BEST ways for the student to

polet [3.4K]

You would have to evaporate the water to get just the sugar

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3 years ago

The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

Learn more: brainly.com/question/4400908

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2 years ago

If the vinegar buret tip were not filled with solution before its initial reading was taken, would the calculated normality of t

Alexxx [7]

The normality that would be calculated will be to high because the change in volume will be greater than the actual change in volume. if the buret tip is not filled when reading the initial volume, the actual volume should be lesser with that reading. so if you will you the higher reading the change in volume or the volume you use in titration will be higher than the actual

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3 years ago

Caustic soda is 19.1 M NaOH and is diluted for household use. What is the household concentration if 10 mL of the concentrated s

olga nikolaevna [1]

Hello!

To calculate the household concentration of NaOH we need to use the dilution formula, clearing for M2, as you can see in the equation below:

$M1*V1=M2*V2 \\ \\ V2= \frac{M1*V1}{V2}$

Now, we input the values from the data we have onto this equation. M1=19,1 M; V1=10 mL; V2=400 mL, and solve the equation to get the result:

$V2= \frac{M1*V1}{V2}= \frac{19,1M*10mL}{400 mL}=0,48M$

So, the household concentration of NaOH will be 0,48 M

Have a nice day!

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3 years ago

What is the probability that an offspring will have a<br> heterozygous genotype? |

Anastasy [175]

Answer:

25,50,25

Explanation:

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2 years ago