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2 years ago

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A pharmacist works with a 1.75 m solution of sodium bromide (nabr) and water. the volume of the solution is 84.0 milliliters. if

the pharmacist dilutes the solution to 1.00 m, what is the volume of the new solution? express your answer to three significant figures. the volume of the new solution is milliliters.

1 answer:

Sav [38]2 years ago

4 0

The volume of the diluted (i.e the new) solution given the data is 147 mL

<h3>Data obtained from the question </h3>

Molarity of stock solution (M₁) = 1.85 M
Volume of stock solution (V₁) = 84 mL
Molarity of diluted solution (M₂) = 1 M
Volume of diluted solution (V₂) = ?

<h3>How to determine the volume of diluted solution </h3>

M₁V₁ = M₂V₂

1.75 × 84 = 1 × V₂

V₂ = 147 mL

Thus, the volume of the new solution is 147 mL

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At which moon position would a person on Earth see the entire half of the moon(full-moon)?

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Sun-earth-moon in a straight line. Earth in the 'middle'.

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3 years ago

Read 2 more answers

Two plane waves of the same frequency and with vibrations in the z-direction are given by c (y, t) = (2 cm) cos a p 4 cm y - 20

My name is Ann [436]

The resultant wave at point (5, 2) is Ψ = 5.99 cos [ 7.15 - (20/s) t]

What are the plane waves:

A plane wave is a special case of wave or field: whose value, at any moment, is constant through any plane that is perpendicular to a fixed direction in space.
plane waves are free-space modes.

Here,

Two plane waves are given:

c (5, t) = 4 cos [(8π/3) - (20/s) t]

c (2, t) = 2 cos [(3π/2) - (20/s) t]

now, the waves as imaginary exponentials,

separating the spatial parts, and then adding them together

we get The resultant:

Ψ = [ 4 sin (8π/3) + 2 sin (3π/2) ]^2 + [ 4 cos(8/3 π) + 2 cos(3/2π) ]^2

Ψ = 5.99 tan(a) = 0.747/ 5.95

a = 7.15

Ψ = 5.99 cos [ 7.15 - (20/s) t]

hence,

The resultant wave is Ψ = 5.99 cos [ 7.15 - (20/s) t]

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Your question is incomplete, but most probably the full question was:

Two plane waves with the same frequency and with vibrations (measured by Psi) in the z-direction are given by c (x, t) = (4cm.) cos [pi/3cm. x - 20/s t + pi] c (y, t) = (2cm.) cos[pi/4cm. y - 20/s t + pi]

Express the waves as imaginary exponentials, separate the spatial parts, and add them together using a phasor diagram to find the resultant at the point x = 5cm. y = 2cm

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2 years ago

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

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4 years ago

A ball is projected horizontally from a table 1.0 m high and hits the ground after falling from a time t=0.45s. If the ball trav

photoshop1234 [79]

The initial velocity is $6.7 m/s$

Explanation:

The motion of the ball is a projectile motion, therefore it consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

In this problem, we just need to analyze the horizontal motion: the horizontal velocity is constant, therefore the horizontal distance travelled is given by

$x=v_x t$

where

$v_x$ is the horizontal velocity

t is the time of flight

Here we have:

t = 0.45 s

x = 3.0 m

And so solving for $v_x$ , we find

$v_x = \frac{3.0}{0.45}=6.7 m/s$

And since the ball was initially projected horizontally, this is also the initial velocity.

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3 years ago

In a horizontal pipe with a gradually decreasing cross-section (in the direction of flow) there is a clear fluid of unknown dens

qwelly [4]

Answer:

the density of this clear fluid is 1140.35 kg/m^3

Explanation:

Given that :

$P_p = P__{Q}} + 260 \ Pa \\ \\ v_p =0.37 \ m/s \\ \\ v_Q = 0.77 \ m/s$

According to Bernoulli's Equation.

$P_p + \frac{1}{2} pv^2_p+pgh_p= P_Q+ \frac{1}{2} pv^2_Q + pgh_Q$

∴ $260 = \frac{1}{2}p (v^2_Q-v^2_p)$ since ( $h_p = h_Q$ )

$\rho = \frac{2(260)}{v^2_Q-v^2_p}$

$\rho = \frac{2(260)}{0.77^2-0.37^2}$

$\rho = \frac{520}{0.5929-0.1369}$

$\rho =$ $1140.35 \ kg/m^3$

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4 years ago