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stiks02 [169]
4 years ago
5

50ml of water is added to a certain amount of solution with a concentration of 0.08g/ml. The final concentration is 0.065 g/ml.

What is the initial volume of water in the solution before the addition was made?
Physics
1 answer:
vladimir1956 [14]4 years ago
7 0

Answer:

<em>216.7 ml</em>

Explanation:

<u>Concentration</u>

The concentration of a substance is defined as the quantity of solute existent in a given quantity of solution. Water is usually a liquid to add to the solute and produce the substance with the required concentration, often expressed in %, gr/ml or any other similar ratio. If the solute is expressed in units of mass and the solution is expressed in units of volume, then the concentration is known as mass concentration or density and is computed as

\displaystyle C=\frac{m_s}{V_t}

Where m_s is the mass of solute and V_t is the total volume of the solution

We know at first  

C_1=0.08\ gr/ml

\displaystyle \frac{m_s}{V_{t1}}=0.08

Operating

m_s=0.08V_{t1}

When we add 50 ml of water and keep the solute unchanged, we have

\displaystyle \frac{m_s}{V_{t1}+50}=0.065

Or equivalently

m_s=0.065(V_{t1}+50)

Equating both masses

0.08V_{t1}=0.065(V_{t1}+50)

0.08V_{t1}=0.065V_{t1}+3.25

Rearranging and simplifying

0.015V_{t1}=3.25

Solving

V_{t1}=216.7\ ml

The mass of the solute is

m_s=0.08V_{t1}=17.3\ gr

We don't know the density of the solute, so we cannot compute its volume which is part of the solution. If we neglect this volume because the concentration is small enough, we can say there were 216.7 ml of water in the initial solution.

If we estimate the density of the solute close to 1 gr/ml, the volume is 17.3 ml, and the initial volume of water in the solution is 216.7-17-3=199.4 ml

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If an object has more protons than electrons, then the net charge on the object is positive. If there are more electrons than protons, then the net charge on the object is negative. If there are equal numbers of protons and electrons, then the object is electrically neutral.

Source: https://www.khanacademy.org/science/ap-physics-1/ap-electric-charge-electric-force-and-voltage/electric-charge-ap/a/electric-charge-ap1

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4 years ago
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Arte-miy333 [17]

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6 0
3 years ago
The velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in
vivado [14]

The force the escaping gas exerts of the rocket is 10.42 N.

<h3>Force escaping gas exerts</h3>

The force the escaping gas exerts of the rocket is calculated as follows;

F = m(v - u)/t

where;

  • m is mass of the rocket
  • v is the final velocity of the rocket
  • u is the initial velocity of the rocket
  • t is time of motion

F = (0.25)(40 - 15)/0.6

F = 10.42 N

Thus, the force the escaping gas exerts of the rocket is 10.42 N.

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7 0
2 years ago
S Problem Set<br> 2.) 6.4 x 109 nm to cm
anyanavicka [17]

Answer:

6.4\cdot 10^2 cm

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

1 nm = 10^{-9} m

So we have:

6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m

Now we can convert from metres to centimetres, keeping in mind that

1 m = 10^2 cm

So, we find:

6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm

8 0
3 years ago
If a car accelerates uniformly from rest to 15 meters
Talja [164]

Answer:

1.125m/s^2

Explanation:

Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.

a = ?

u = 0m/s

v = 15m/s

s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

225= 200a

Divide both sides by 200

225/200 = 200a/200

a= 1.125m/s^2

Hence the acceleration of the car is 1.125m/s^2.

Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s

8 0
3 years ago
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