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Rina8888 [55]
2 years ago
5

Which situation is the best example of regulation in an economic system?

Chemistry
1 answer:
velikii [3]2 years ago
5 0

Hope this helps!

Answer: The correct option is: "A state agency has been created to monitor the production and distribution of sports drinks."

Explanation:

The economic regulation are the provisions through which the government intervenes in the markets to set prices or quantities of production, or establish technical specifications and in general, restrictions that must be met by citizens and companies to participate in a market.

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When a 2.00 g sample of KCl is dissolved in water in a calorimeter that has a total heat capacity of 1.28 kJ ⋅ K − 1 , the tempe
Alchen [17]

Answer : The molar heat of solution of KCl is, 17.19 kJ/mol

Explanation :

First we have to calculate the heat of solution.

q=c\times (\Delta T)

where,

q = heat produced = ?

c = specific heat capacity of water = 1.28kJ/K

\Delta T = change in temperature = 0.360 K

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

q=1.28kJ/K\times 0.360K

q=0.4608kJ=460.8J

Now we have to calculate the molar heat solution of KCl.

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

where,

\Delta H = enthalpy change = ?

q = heat released = 460.8 J

m = mass of KCl = 2.00 g

Molar mass of KCl = 74.55 g/mol

\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{2.00g}{74.55g/mole}=0.0268mole

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

\Delta H=\frac{460.8J}{0.0268mole}

\Delta H=17194.029J/mol=17.19kJ/mol

Therefore, the molar heat of solution of KCl is, 17.19 kJ/mol

7 0
3 years ago
Calculate the vapor pressure of water at T=90°C
LenKa [72]

The equilibrium vapour pressure is typically the pressure exerted by a liquid .... it is A FUNCTION of temperature...

Explanation:

By way of example, chemists and physicists habitually use

P

saturated vapour pressure

...where

P

SVP

is the vapour pressure exerted by liquid water. At

100

∘

C

,

P

SVP

=

1

⋅

a

t

m

. Why?

Well, because this is the normal boiling point of water: i.e. the conditions of pressure (i.e. here

1

⋅

a

t

m

) and temperature, here

100

∘

C

, at which the VAPOUR PRESSURE of the liquid is ONE ATMOSPHERE...and bubbles of vapour form directly in the liquid. As an undergraduate you should commit this definition, or your text definition, to memory...

At lower temperatures, water exerts a much lower vapour pressure...but these should often be used in calculations...especially when a gas is collected by water displacement. Tables of

saturated vapour pressure

are available.

4 0
3 years ago
Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL
dybincka [34]

Answer:

M_2=3.34x10^{-3}M

Explanation:

Hello!

In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

M_1V_1=M_2V_2

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M

Best regards!

5 0
3 years ago
Why are teeth not minerals?
lozanna [386]
Because your Teeth are composed of calcium, phosphorus, and other minerals. ... But bones are still not as strong as teeth. The hardest part of the human body ,teeth mostly consist of a calcified tissue called dentine. The tooth's dentine tissue is covered in enamel, that hard, shiny layer that you brush.


7 0
3 years ago
Read 2 more answers
A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

8 0
3 years ago
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