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3 years ago

Why are suspensions and colloids opposites concerning the way the particles in them behave?

2 answers:

8 0

Answer: The particles in a suspension may be separated by filtration unlike those in a colloid. Colloids are able to scatter light, but suspensions cannot transmit light. Particles in a suspension can be seen by the naked eye, but those in a …

Explanation:

bearhunter [10]3 years ago

7 0

Answer:

Im pretty sure they bahave differently because The particles in a suspension may be separated by filtration unlike those in a colloid. Colloids are able to scatter light, but suspensions cannot transmit light. Particles in a suspension

Explanation:

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Inessa05 [86]

B. Temporary hope this helped :)

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2 years ago

If it takes 22974 J to heat 96.5 g of a substance from 13.46°C to 73.86 °C, what is the specific heat of the substance?

yulyashka [42]

The specific heat capacity, C, of the substance is 3.94 $\frac{J}{g-^oC}$ . To raise one gram of a substance by one degree Celsius, a specific amount of heat must be absorbed. This is called the specific heat of the substance. The higher the specific heat of a substance, the more energy is needed to increase its temperature and vice versa.

Further Explanation:

The specific heat capacity, C, of a substance is related to the amount of substance and the temperature change using the following equation

$q \ = \ mC(T_{f}-T_{i})$

where

q = amount of heat absorbed (usually in J)

m = amount of substance (in g)

C = specific heat capacity (in $\frac{J}{g-^oC}$ )

T(f) = final temperature

T(i) = initial temperature.

To answer the problem, we sort the given information:

q = 22974 J (5 significan figures)

m = 96.5 g (3 significant figures)

C = ?

T(f) = 73.86 °C (4 significant figures)

T(i) = 13.46 °C (4 significant figures)

Then, to find C, rearrange the equation:

$C \ = \frac{q}{m(T_{f} \ - \ T_{i})}$

Plug in the values to the equation above to get the value of C.

$C \ = \ \frac{22974 \ J}{96.5 \ g \ (73.86 \ \ -13.46 \°C)} \\C \ = \ \frac{22974 \ J}{96.5 \ g \ (60.4\°C)} \\C \ = \ \frac{22974 \ J}{5828.6 \ g -\°C} \\C \ = \ 3.941598 \ \frac{J}{g -\°C} \\$

Since the least number of significant figures in the given is 3, then the final answer must have 3 significant figures, too. Therefore,

$\boxed {C = 3.94 \ \frac{J}{ g -\°C}}$

Learn More:

1. Learn more about heat capacity brainly.com/question/8828503

2. Learn more about calorimetry brainly.com/question/8168263

3. Learn more about heat of reaction brainly.com/question/10122365

Keywords: heat capacity, specific heat, calorimetry

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3 years ago

How long do it take for copper to form

kondor19780726 [428]

Copper is an element. So is can't form, but it takes certain types of copper tens of thousands of years depending on where the magma was cooled.

6 0

3 years ago

Read 2 more answers

What volume of 0.305 m agno3 is required to react exactly with 155.0 ml of 0.274 m na2so4 solution? hint: you will want to write

LuckyWell [14K]

The balanced chemical equation for reaction of $AgNO_{3}$ and $Na_{2}SO_{4}$ is as follows:

$2 AgNO_{3}+Na_{2}SO_{4}\rightarrow 2NaNO_{3}+Ag_{2}SO_{4}$

From the balanced chemical equation, 2 mol of $AgNO_{3}$ reacts with 1 mol of $NaNO_{3}$ .

First calculating number of moles of $NaNO_{3}$ as follows:

$M=\frac{n}{V}$

On rearranging,

$n=M\times V$

Here, M is molarity and V is volume. The molarity of $NaNO_{3}$ is given 0.274 M or mol/L and volume 155 mL, putting the values,

$n=0.274 mol/L\times 155\times 10^{-3}mL=0.04247 mol$

Since, 1 mol of $NaNO_{3}$ reacts with 2 mol of $AgNO_{3}$ thus, number of moles of $AgNO_{3}$ will be $2\times 0.04247 mol=0.08494 mol$ .

Now, molarity of $AgNO_{3}$ is given 0.305 M or mol/L thus, volume can be calculated as follows:

$V=\frac{n}{M}=\frac{0.08494 mol}{0.305 mol/L}=0.2785 L=278.5 mL$

Therefore, volume of $AgNO_{3}$ is 278.5 mL.

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3 years ago

What is the compound name for NF

dlinn [17]

Nitrogen trifluoride is the inorganic compound with the formula NF3. This nitrogen-fluorine compound is a colorless, odorless, nonflammable gas. It finds increasing use as an etchant in microelectronics.

(If you mean by a gas of some sort)

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2 years ago

Read 2 more answers