Which letter pattern best represents seafloor spreading? (Hint: Letter "f" represents the mid-

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

6. An alloy of iron contains 75.0% iron and 25.0% other elements. How many grams of iron are present in 150. g of the alloy?

Hoochie [10]

Answer:

108.5

Explanation:

half of 150 is 75 and half of that is 32.5. Add that to 75 and you'll get your answer

3 0

3 years ago

What volume of 0. 250 m sulfuric acid solution would be needed to react completely with 18. 00 ml of 0. 350 m ammonia solution?

PolarNik [594]

The volume of 0. 250 m sulfuric acid solution would be needed to react completely with 18. 00 ml of 0. 350 m ammonia solution is 0.0504 L.

The balance reaction is 2KOH+ $H_{2} SO_{4}$ = $2H_{2} O+K_{2} SO_{4}$

This shows us that we need 2 moles of KOH for each mole of sulfuric acid.

So, we write mathematically 2*(0.018*0.350)=(V*0.250)

V=(2*0.018*0.350)/0.250=0.0504 L

<h3>Sulfuric acid</h3>

With the chemical formula H2SO4, sulfuric acid, also known as sulphuric acid or oil of vitriol, is a mineral acid made up of the elements hydrogen, oxygen, and sulfur. It is a viscous liquid that is miscible with water and is colorless, odorless, and viscous.

Since pure sulfuric acid has a significant affinity for water vapor, it does not naturally occur on Earth; as a result, it is hygroscopic and rapidly collects water vapor from the atmosphere. Due to its potent dehydrating and oxidizing qualities, concentrated sulfuric acid is extremely corrosive to various materials, including rocks and metals. One significant exception is phosphorus pentoxide, which dehydrates sulfuric acid to sulfur trioxide instead of being dehydrated by sulfuric acid.

Learn more about Sulfuric acid here:

brainly.com/question/1107054

#SPJ4

6 0

2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

3 years ago

What is ejected from the nucleus during alpha decay?

Aloiza [94]

Hi

Alpha decay is a type of radioactive decay.
So, when Alpha decay goes through a process called "radioactive process", two neutrons and two protons comes together and is ejected from the nucleus of a radioactive atom.

Hope this helped :)

7 0

3 years ago