Taking into account the definition of molarity, the concentration of a solution that contains 70 g of H₂SO₄ in 0,28 dm³ of solution is 2.5510 $\frac{moles}{L}$ .

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

$Molarity=\frac{number of moles}{volume}$

Molarity is expressed in units $\frac{moles}{L}$ .

In this case, you have:

number of moles= 70 g× $\frac{1 mole}{98 g}$ = 0.7143 moles, where 98 g/mole os the molar mass of H₂SO₄
volume= 0.28 dm³= 0.28 L (being 1 dm³= 1 L)

Replacing in the definition of molarity:

$Molarity=\frac{0.7143 moles}{0.28 L}$

Solving:

<u><em>Molarity= 2.5510 </em></u> $\frac{moles}{L}$

Finally, the concentration of a solution that contains 70 g of H₂SO₄ in 0,28 dm³ of solution is 2.5510 $\frac{moles}{L}$ .

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4 0

1 year ago

A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w

tatyana61 [14]

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s$
And so, the kinetic energy of the object is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s$
And so the new kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

7 0

2 years ago