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3 years ago

How would you prepare 2.5 L of a 0.800M solution of KNO3?

1 answer:

7 0

First find the mass of solute:

Molar mass KNO₃ = 101.1032 g/mol

mass = Molarity * molar mass * volume

mass = 0.800 * 101.1032 * 2.5

mass = 202.2064 g of KNO₃

To prepare 2.5 L (0800 M) of KNO3 solution, must weigh 202.2064 g of salt, dissolve in a Beker, transfer with the help of a funnel of transfer to a volumetric flask, complete with water up to the mark, capping the balloon and finally shake the solution to mix.

hope this helps!

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Is hot tea homogeneous or heterogeneous?

OleMash [197]

homogeneous because particles are mixed uniformly

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3 years ago

There are several aromatic compounds with the formula C8H9Cl. Draw those that have a trisubstituted ring where the methyl groups

Kryger [21]

Answer:

$3-Chloro-1,2-dimethylbenzene$

$4-Chloro-1,2-dimethylbenzene$

Explanation:

Such types of compounds in which conjugated planer ring system and delocalized pi electrons are present are called aromatic compounds such as Toluene, Benzene and some other. Generally, these compounds follow Huckle's rule. The trisubstituted ring means a compound in which three hydrogen atoms are replaced by three other groups. In the given compound two hydrogen atoms are replaced by two methyl groups and one hydrogen atom is replaced by one Chlorine atom. From the given compound $3-Chloro-1,2-dimethylbenzene$ , and $4-Chloro-1,2-dimethylbenzene$ can be drawn.

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3 years ago

Which of these elements is the most reactive? potassium calcium titanium scandium

vladimir1956 [14]

Answer:

Scandium is the most reactive

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3 years ago

A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at

Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $heat_{absorbed}=heat_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 45 g

$m_2$ = mass of coffee = 180 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $85^oC$

$c_1$ = specific heat of aluminium = $0.80J/g^oC$

$c_2$ = specific heat of coffee= $4.186 J/g^oC$

Putting all the values in equation 1, we get:

$45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]$

$T_{final}=80.30^oC$

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

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3 years ago

How can forgery be used in crime

Tems11 [23]

Some bad people will tend to forge dollar bills and make fake bills, giving them to cashiers and asking for it to be split into a different amount of bills so they are given real cash and wont be caught or they will buy things, return them and be given back real money and get away with it. This is obviously a crime because it's fake.

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3 years ago

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