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3 years ago

A 6.0 Liter solution is made with 2 moles of solute. What is the molarity of the solution?

Chemistry

1 answer:

kodGreya [7K]3 years ago

3 0

The formula for molarity is: mol/L. so it would be 2.0 mol/ 6.0 L.

the answer is: .33 M

hope this helps!

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State Modern periodic Table

Kaylis [27]

Answer:

According to the modern periodic law, the properties of the elements and their compounds are a periodic function of their atomic numbers. Thus, in the modern periodic table, atomic number forms the basis of the classification of elements;The modern table is called 'long form' of the periodic table.

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3 years ago

12. help idk what to do at this point , I have notes and study till 3am , I still need help

GaryK [48]

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2 years ago

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Activity: A Ferris wheel with a diameter of 60.0 m is moving at a speed of 2.09 m/s. What is the centripetal

GarryVolchara [31]

Answer:

Centripetal acceleration of the wheel is $0.145\ m/s^2$ .

Explanation:

We have,

Diameter of a Ferris wheel is 60 m

Radius of the wheel is 30 m

Speed of the wheel is 2.09 m/s

It is required to find the centripetal acceleration of the wheel. The formula of centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(2.09)^2}{30}\\\\a=0.145\ m/s^2$

So, the centripetal acceleration of the wheel is $0.145\ m/s^2$ .

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2 years ago

How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)

Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $\frac{100 g}{44 g/mol}=2.273 moles$

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = $2.273\times 10^{-3} kmol$

2) 1 liter of ethyl alcohol of density $0.789 g/cm^3$

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = $0.789 g/cm^3$

$1 cm^3=1 mL$

Mass of ethyl alcohol = m

$m=d\times V=0.789 g/cm^3\times 1000 mL=789 g$

Molar mass of ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = $\frac{789 g}{46 g/mol}=17.152 mol$

$17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol$

3) Volume of oxygen gas,V = $1.5 m^3=1500 L$

$1 m^3= 1000 L$

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

$PV=nRT$

$n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol$

0.01632 mol = 0.01632 × 0.001 kmol= $1.632\times 10^{-5} kmol$

4) Volume water in mixture = 1 L

Density of water = $1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L$

Mass of water = $1000 g/L\times 1 L = 1000 g$

Volume of alcohol = 2.5 L

Density of alcohol = $789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L$

Mass of alcohol = $789 g/L\times 2.5 L = 1972.5 g$

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

$\frac{1000 g}{2972.5 g}\times 100=33.64\%$

Mass percentage of alcohol :

$\frac{1972.5 g}{2972.5 g}\times 100=66.36\%$

Moles of water :

$n_1=\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of alcohol =

$n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol$

Mole fraction of water :

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644$

Mole fraction of alcohol :

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356$

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2 years ago

A population of mice, some with light-colored fur and some with dark-colored fur are introduced into a field with dark soil. A f

Iteru [2.4K]

Answer:

Dark-colored mice are better able to hide from predators.

Explanation:

If all of the light mice are eaten, then there are none left to pass on the light-furred gene, meaning that there would be no offspring without dark fur.

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2 years ago

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