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Tju [1.3M]
2 years ago
10

(sand) and (Sand with water) both of them are heterogeneous mixture isn't it?​

Chemistry
1 answer:
patriot [66]2 years ago
8 0

Answer:

yes true

Explanation:

both are heterogeneous

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Which of the following elements can form diatomic molecules held together by triple covalent bonds?
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Nitrogen can form a diatomic molecule held together by triple bonds.
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Rank the following from lowest to highest anticipated boiling point: C8H18, CH, C16H34,CH4,C4H10
viva [34]
<span>The relative strength of intermolecular forces such as ionic, hydrogen bonding, dipole-dipole interaction and Vander Waals dispersion force affects the boiling point of a compound. For this case, the longer the chain the higher the boiling point. 

</span>CH, CH4, C4H10, C8H18, C16H34

Hope this answers the question. Have a nice day.
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3 years ago
Convert 59,800 cg/L to g/mL
alexgriva [62]
To convert the given value to the desired one, use the proper unit conversions and dimensional analysis. Use the following conversion for the first set.

    1 g = 100 cg
    1 L = 1000 mL

Using the concept presented above,
         
            V = (59800 cg/L)(1 g/100 cg)1 L/1000 mL)
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6 0
3 years ago
1. Currently, California students are in class 240 minutes every day. How many hours is this?
sweet [91]

Answer:

4 hours a day

Explanation:

240/60=4

3 0
3 years ago
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Propane gas, C3H8, is sometimes used as a fuel. In order to measure its energy output as a fuel a 1.860 g sample was combined wi
lisov135 [29]

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ

Moles of propane =\frac{1.860 g}{44 g/mol}=0.0422 mol

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol

3 0
3 years ago
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