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pishuonlain [190]
3 years ago
8

Sugar cannot be separated from sugar solution by filtering explain why?​

Chemistry
2 answers:
luda_lava [24]3 years ago
7 0

Answer:

Sugar can not be separated due it has very fine particles.

Explanation:

When we use to separate sugar with salt they both are dissolved with the water.

alexdok [17]3 years ago
5 0

Answer:

The sugar particles are so small that they have dissolved in the water, and can easily pass through the filter.

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What is the name of this compound?<br><br> 2-butanone<br> 2-pentanone<br> 4-pentanone<br> 4-butanone
UkoKoshka [18]

Answer:

2-pentanone

Explanation:

The oxygen atom is attached to the 2nd carbon atom in the longest linear chain, thus the prefix 2. the longest continuous carbon chain has five carbon atoms giving it the prefix pent-. the chemical is a ketone thus the suffix none.

7 0
4 years ago
Read 2 more answers
a container with a volume of 3.46 liters contains a gas at a pressure of 1.30 atm and temperature of 274.27 kelvins how many mol
balu736 [363]
There are 0.19 ((1.3*3.46)/(0.082*274.27) moles of gas are in the container. This problem can be solved using the ideal gas law formula which stated as PV=nRT where P is the gas' pressure, v is the gas' volume, n is the gas' mole amount, R is the constant gas standard (0.082), and T is the gas' temperature<span>.</span>
8 0
3 years ago
1-Bromopropane is treated with each of the following reagents. Draw the major substitution product if the reaction proceeds in g
sammy [17]

Answer:

Explanation:

If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane  a primary alkyl-halide.

Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used  otherwise it will result to a reactive mechanism if a weak electrophile is used.

However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.

The attached diagrams portraying this notions is shown in the attached file below.

5 0
3 years ago
Could someone please care to explain what the answer is
polet [3.4K]

Calculate the percentage composition of Ca(MnO4)2.

<u>Firstly, we see how many atoms are there for each element in the formula</u>.

Ca= 1 atom

Mn= 2 atoms

O= 8 atoms

<u>Next, we are going to consult our periodic table for the atomic mass of each element.</u>

Ca= 40

Mn= 55

O= 16

Then, we have to find the molar mass for the compound..

Here is the formula for calculating molar mass of an element:

Molar Mass= ( no. of atoms of the element × atomic mass of the element)

Now, we have to calculate the atomic mass of the compound. So using the molar mass formula for an element, we calculate the molar mass for each element then we sum up their molar masses to get the compounds molar mass.

Molar mass (Ca)= 1× 40

(Ca)= 40

Molar mass (Mn)=2×55

(Mn)= 110

Molar mass (O)= 8×16

(O)= 128

Now: Molar mass( compound)= (Ca)+(Mn)+(O)

= 40+ 110 128

= 278

This is everything we need to calculate our percentage composition for each element..

* The example says to find the percentage composition for Ca. So we only find for Ca, Which is already done using the formula and the answer is 14.39%.

To prove that your answer is correct, find the percentage composition for Mn and O as well. Then you add up their percentage compositions.

If you do and you get 100 as your answer, then your percentage compositions are correct.

Why don't you try finding the percentage composition for Mn and O, then add up all the three percentage compositions. If you 100 as their sum, then your percentage composition for each of the elements are correct.

6 0
3 years ago
What is the quantity of heat (in kJ) associated with cooling 185.5 g of water from 25.60°C to ice at -10.70°C?Heat Capacity of S
Cerrena [4.2K]

Taking into account the definition of calorimetry, sensible heat and latent heat,  the amount of heat required is 37.88 kJ.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

<h3>Latent heat</h3>

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

  • <u><em>25.60 °C to 0 °C</em></u>

First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.

So, you must lower the temperature from 25.60°C (in liquid state) to 0°C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • c= Heat Capacity of Liquid= 4.184 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C

Replacing:

Q1= 4.184 \frac{J}{gC}× 185.5 g× (- 25.6 °C)

Solving:

<u><em>Q1= -19,868.98 J</em></u>

  • <u><em>Change of state</em></u>

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

n= 185.5 grams× \frac{1mol}{18 grams}= 10.30 moles, where 18 \frac{g}{mol} is the molar mass of water, that is, the amount of mass that a substance contains in one mole.

ΔHfus= 6.01 \frac{kJ}{mol}

Replacing:

Q2= 10.30 moles×6.01 \frac{kJ}{mol}

Solving:

<u><em>Q2=61.903 kJ= 61,903 J</em></u>

  • <u><em>0 °C to -10.70 °C</em></u>

Similar to sensible heat previously calculated, you know:

  • c = Heat Capacity of Solid = 2.092 \frac{J}{gC}
  • m= 185.5 g
  • ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C

Replacing:

Q3= 2.092 \frac{J}{gC} × 185.5 g× (-10.70) °C

Solving:

<u><em>Q3= -4,152.3062 J</em></u>

<h3>Total heat required</h3>

The total heat required is calculated as:  

Total heat required= Q1 + Q2 +Q3

Total heat required=-19,868.98 J + 61,903 J -4,152.3062 J

<u><em>Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ</em></u>

In summary, the amount of heat required is 37.88 kJ.

Learn more about calorimetry:

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7 0
2 years ago
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