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jarptica [38.1K]

3 years ago

7

You discover a new substance that mixes well with H2O initially, but eventually the particles sink to the bottom. What is the BE

ST term to describe this mixture?
A.
Colloid

B.
Solution

C.
Homogeneous

D.
Suspension

1 answer:

AVprozaik [17]3 years ago

7 0

I believe this would be colloid

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3 years ago

9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r

Mumz [18]

Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

$\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O$

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

$\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\ \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\ \\ Percent\text{ }yield=0.9484\times100\% \\ \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}$

Therefore, the percent yield for the reaction is 95%.

3 0

1 year ago

What is the total number of electron pairs shared between the carbon atom and one of the oxygen atoms in a carbon dioxide molecu

TEA [102]

The answer is 2 pairs. The carbon atom has 4 electrons and oxygen atom has 6 electrons. The purpose of shared electron is to get 8 electrons at the outer layer. So the carbon atom needs to share 4 electrons which is 2 pairs with oxygen to be stable.

7 0

3 years ago

If the pressure on a gas at -73°C is doubled but its volume is held constant, what will its final temperature be in degrees Cels

Gre4nikov [31]

Answer:

127°C

Explanation:

This excersise can be solved, with the Charles Gay Lussac law, where the pressure of the gas is modified according to absolute T°.

We convert our value to K → -73°C + 273 = 200 K

The moles are the same, and the volume is also the same:

P₁ / T₁ = P₂ / T₂

But the pressure is doubled so: P₁ / T₁ = 2P₁ / T₂

P₁ / 200K = 2P₁ / T₂

1 /2OOK = (2P₁ / T₂) / P₁

See how's P₁ term is cancelled.

200K⁻¹ = 2/ T₂

T₂ = 2 / 200K⁻¹ → 400K

We convert the T° to C → 400 K - 273 = 127°C

4 0

3 years ago