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Olegator [25]
3 years ago
7

Help with this chemistry question please???

Chemistry
1 answer:
givi [52]3 years ago
6 0
To find percent composition, you should first find the total of all of the components. Just add them all together:

100g+200g+50g=350g

Then, divide the amount of the component you are interested in, in this case, chocolate (50g), by the total amount to give:

\frac{50g}{350g}

This will give you the answer in decimal form, but to convert to a percent, you need to multiply by 100:

\frac{50g}{350g}*100

Therefore, the answer choice that is correct is A.
You might be interested in
Look at the following enthalpy diagram. Select all that apply.
Drupady [299]

Answer:

Option 2 and 4 are correct

Explanation:

The reactants in the attached image have more enthalpy and hence less stability as they are more reactive. Thus, Product is more stable than the reactants.

This is an addition reaction in which two reactants add up to form the product.

Very less activation energy is required as the reactants themselves are unstable, possess high energy and hence are very reactive.

Reactants have more energy than the products.  

5 0
3 years ago
A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to th
Yuki888 [10]

The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.  

The mass percent of lithium hydroxide can be calculated with the following equation:  

\% = \frac{m_{LiOH}}{m_{t}} \times 100    (1)

Where:

m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g   (2)  

We need to find the mass of LiOH.

From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.    

\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}

0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}

\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    (3)                                        

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:  

\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol    

\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol              

Solving for m_{LiOH}, we have:

m_{LiOH} = 0.082 g

Hence, the percent lithium hydroxide is (eq 1):

\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%  

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

Learn more about mass percent here:

  • brainly.com/question/6992535?referrer=searchResults
  • brainly.com/question/5840377?referrer=searchResults

I hope it helps you!                        

5 0
2 years ago
Which of the following statements is true?
Anna007 [38]

Answer:

C. Lithium is most easily oxidized of the metals listed on the activity series and therefore it will most easily give electrons to metal cations

Explanation:

"Lithium" is a type of alkali metal that has a "single valence electron." Since it is a reactive element, it easily gives up an electron when it is combined with other elements. Such giving up of electron is meant to create compounds or bonds.

Among the common metals listed, "lithium" is the most easily oxidized. This means that it donates its electrons immediately. Such combination makes it exist as a<em> "cation"</em> or <em>"positively-charged."</em>

So, this explains the answer.

6 0
3 years ago
PLEASE HELP When I combine Sprite with a sour candy, it starts to bubble a lot. Is this a physical or chemical change?
maksim [4K]

Answer:

B. chemical

Explanation:

Chemical change cannot go back to its original form

4 0
2 years ago
Read 2 more answers
a gas occupies 600.0 ml at 20.00 degrees C and 70.00kPa what will be the pressure at 40.00 degrees C and 150.0mL?
Zina [86]

Answer:

P₂ = 299.11 KPa

Explanation:

Given data:

Initial volume = 600 mL

Initial pressure = 70.00 KPa

Initial temperature = 20 °C (20 +273 = 293 K)

Final temperature = 40°C (40+273 = 313 K)

Final volume = 150.0 mL

Final pressure = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₂ = P₁V₁ T₂/ T₁ V₂  

P₂ =  70 KPa × 600 mL × 313 K / 293K ×150 mL  

P₂ = 13146000 KPa .mL. K /43950 K.mL

P₂ = 299.11 KPa

4 0
3 years ago
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