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vekshin1
3 years ago
5

Choose the options below that are true of a solution of a solid in a liquid. (select all that apply) Select all that apply: Most

solids in solution exhibit a general trend of increasing solubility with increasing temperature. A seed crystal may be added to a supersaturated solution to precipitate excess solute. Most solids in solution will dissolve less with increasing temperature. A solution can be saturated at an elevated temperature and subsequently cooled to a lower temperature without precipitating the solute.
Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
6 0

Answer:

Most solids in solution exhibit a general trend of increasing solubility with increasing temperature.

A seed crystal may be added to a supersaturated solution to precipitate excess solute.

Explanation:

For many solids dissolved in liquid water, the solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions(Lumen Learning).

When a seed crystal is added to a supersaturated solution, excess solute begin to precipitate because  the seed crystal now furnishes the required nucleation site where the excess dissolved crystals now begin to grow.

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Determine the number of moles of hydrogen atoms in each of the following.
pashok25 [27]

Answer:

0.855molH

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to realize this problem is about mole ratios. Thus, for the compound C4H10, note there is a 1:10 mole ratio to the hydrogen atoms, that is why the number of moles of the latter is calculated as shown below:

8.55x10^{-2} mol C_4H_{10}*\frac{10molH}{1molC_4H_{10}} \\\\=0.855molH

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3 0
3 years ago
Can some one help I'm lost iam being timed:( Show using two conversion factors how you would convert from 0.020 kg into mg
dezoksy [38]

Answer:

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Explanation:

5 0
3 years ago
The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

7 0
3 years ago
An ice cube at 0.00C with a mass of 8.32g is placed Into 55g of water, initially at 25C. If no heat is lost to the surroundings,
devlian [24]

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Thats all i know

7 0
3 years ago
What is the electron configuration for oxygen with a 2- charge (o2-)?
Juli2301 [7.4K]
Oxygen:
Atomic no. = 8(from periodic table)
⇒1s^2 2s^2 2p^4
But it is O^2-
There are 2 more electrons
=>1s^2 2s^2 2p^6
Voila!
3 0
3 years ago
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