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jeka57 [31]
4 years ago
14

What is the molarity of a solution that contains 6 moles of solute in 2 liters of solution?

Physics
1 answer:
ankoles [38]4 years ago
5 0

Answer:

3mol/L or 3mol/dm³

Explanation:

Given:

solution contains 6 moles of solute consider it 'n' i.e 6 moles

Volume of solution consider in 'V' i.e 2 liters

Required:

Molarity of solution 'M'=?

Formula:

In order to find  Molarity of a solution, no.of moles of solute divided by Vol.of solution in litres (∴ number of moles in 1 litre of solution)

Molarity = no.of moles of solute/Vol.of solution in litres or dm³

Calculations:

M= 6mol/2L =

M= 3mol/L or 3mol/dm³

Therefore, the molarity of a solution that contains 6 moles of solute in 2 liters of solution is 3mol/L or 3mol/dm³

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4 years ago
Explain why earth's gravity pulls you toward the ground
pogonyaev

Answer:

Earth's gravity comes from all its mass. All its mass makes a consolidated gravitational draw on all the mass in your body.

Explanation:

5 0
3 years ago
Determine the speed of sound in air at 300 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3
amm1812

Answer:

c_3_0_0_K=347.19m/s

M=0.864

Explanation:

The speed of sound in the air increases 0.6 m / s for every 1 ° C increase in temperature.  An approximate speed can be calculated using the following empirical formula:

c=331.5+0.6\vartheta

Where:

\vartheta=T-273.15K\\\\

A more exact equation, usually referred to as adiabatic velocity of sound, is given by the following formula:

c=\sqrt{k*R*T}

Where:

R= Gas\hspace{3}constant\hspace{3}of\hspace{3}air=0.287kJ/kg*K=287J/kg*K\\k=Specific\hspace{3}heat\hspace{3}ratio=1.4\\T=Temperature=300K

Hence:

c=\sqrt{(287)*(1.4)*(300)} =347.1887095\approx347.19m/s

Now, the Mach number at which an aircraft is flying can be calculated by:

M=\frac{u}{c}

Where:

u= Velocity\hspace{3}of\hspace{3}the\hspace{3}moving\hspace{3}aircraft\\c= Speed\hspace{3}of\hspace{3}sound\hspace{3}at\hspace{3}the\hspace{3}given \hspace{3}altitude

Therefore:

M=\frac{300}{347.19} =0.8640833984\approx0.864

5 0
4 years ago
A 18-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 30 N. Starting from rest, the sle
k0ka [10]

Answer:

Coefficient of kinetic friction = 0.146

Explanation:

Given:

Mass of sled (m) = 18 kg

Horizontal force (F) = 30 N

FInal speed (v) = 2 m/s

Distance (s) = 8.5 m

Find:

Coefficient of kinetic friction.

Computation:

Initial speed (u) = 0 m/s

v² - u² = 2as

2(8.5)a = 2² - 0²

a = 0.2352 m/s²

Nweton's law of :

F (net) = ma

30N - μf = 18 (0.2352)

30 - 4.2336 = μ(mg)

25.7664 =  μ(18)(9.8)

μ = 0.146

Coefficient of kinetic friction = 0.146

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