1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mojhsa [17]
3 years ago
15

A student is told to use 20.0 g of sodium chloride to make an aqueous solution that has a concentration of 10.0 g/L (grams of so

dium chloride per liter of solution). Assuming that 20.0 g of sodium chloride has a volume of 7.50 mL, show that she will need about 1.99 L of water to make this solution. In making this solution, should she add the solute to the solvent or the solvent to the solute?
Physics
2 answers:
Stells [14]3 years ago
6 0

Answer:

she should add solute to the solvent

Explanation:

Given data :

Mass of the sodium chloride, = 20.0 g

Concentration of the solution = 10 g/L

Volume of 20.0 g of sodium chloride = 7.50 mL

Now, from the concentration, we can conclude that for 10 g of sodium chloride volume of the solution is 1 L

thus, for 20 g of sodium chloride  volume of the solution is 2 L or 2000 mL

also,

Volume of solution = Volume of solute(sodium chloride) + volume of solvent (water)

thus,

2000 mL = 7.5 mL + volume of solvent (water)

or

volume of water = (2000 - 7.5) mL

or

volume of water = 1992.5 mL

or

volume of water = 199.25 L ≈ 199 L

3241004551 [841]3 years ago
5 0

Explanation:

Concentration of sodium chloride = 10.0 g/L

1 Liters of solution has 10 g of sodium chloride

The volume of solution in which 20 g of sodium chloride will present is:

\frac{1 L}{10 }\times 20 =2 L

Volume of solution we will be making is of 2 L.

Volume of 20 g of sodium chloride = 7.50 ml = 0.00750 L

Volume of the water = ?

Volume of solution = Volume of solute + Volume of solvent

Volume of solvent = 2L - 0.00750 L = 1.9925 L

So,volume of water required to prepare the 2 L of 10 g/L solution of sodium chloride is 1.9925 L.

And for while making of the solution , student must add solute the solvent.

You might be interested in
An example of a high energy electromagnetic wave is
Wewaii [24]
<span>An example of a high energy electromagnetic wave is "X-Ray"

When car runs, it's chemical energy (gasoline) converts into mechanical energy

Temperature is the measure of hotness or coldness of the body, so when heat expose to a substance, it's degree of hotness increases & it's temperature increases

Hope this helps!
</span>
4 0
3 years ago
A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.
LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person  standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = \frac{MR^2}{2}

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia (I_y_y )  of the system with respect to the axis yy.

⇒ I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g

⇒ I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2

⇒ I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2

⇒ I_y_y=289.088\ kg.m^2

The moment of inertia of the system is 289.088 kg.m^2

7 0
3 years ago
A 60.0 kg girl stands up on a stationary floating raft and decides to go into shore. She dives off the 180 kg floating raft with
padilas [110]
Momentum, p = m.v
m of the girl = 60.0 kg
m of the boat = 180 kg
v of the girl = 4.0 m/s

A) Momentum of the girl as she is diving:
p = m.v = 60.0 kg * 4.0 m/s = 24.0 N/s

B) momentum of the raft = - momentum of the girl = -24.0 N/s

C) speed of the raft

p = m.v ; v = p/m = 24.0N/s / 180 kg = -0.13 m/s [i.e. in the opposite direction of the girl's velocity]
8 0
3 years ago
The waste products of a nuclear fission powerplant can best be described as
OverLord2011 [107]
A.
 
-> the products are stable atoms with a much lower atomic number. For example, Uranium 238 commonly breaks down into strontium and xenon, both of which are stable and non-radioactive.

They are also low in quantity because of the precision involved in breaking down a large atom via neutron capture means that few fission reactions take place.

Hope I helped :)

Source: I studied nuclear physics in year 11
4 0
3 years ago
Read 2 more answers
Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates
olasank [31]

2) 20.2 m/s

In the first 4.4 seconds of its motion, the blue car accelerates at a rate of

a=4.6 m/s^2

So its final velocity after these 4.4 seconds is

v=u+at

where

u = 0 is the initial velocity (the car starts from rest)

a is the acceleration

t is the time

Substituting t = 4.4 s, we find

v=0+(4.6)(4.4)=20.2 m/s

After this, the car continues at a constant speed for another 8.5 s, so it will keep this velocity until

t=4.4 + 8.5 = 12.9 s

Therefore, the velocity of the car 10.4 seconds after it starts will still be 20.2 m/s.

3) 216.2 m

The distance travelled by the car during the first 4.4 s of the motion is given by

d_1 = ut_1 + \frac{1}{2}at_1^2

where

u = 0 is the initial velocity

t_1 = 4.4 s is the time

a=4.6 m/s^2 is the acceleration

Substituting,

d_1 = 0 +\frac{1}{2}(4.6)(4.4)^2=44.5 m

The car then continues for another 8.5 s at a constant speed, so the distance travelled in the second part is

d_2 = vt_2

where

v_2 = 20.2 m/s is the new velocity

t_2 = 8.5 s is the time

Substituting,

d_2 = (20.2)(8.5)=171.7 m

So the total distance travelled before the brakes are applied is

d=44.5 m+171.7 m=216.2 m

4) -6.62 m/s^2

We are told that the blue car comes to a spot at a distance of 247 meters from the start. Therefore, the distance travelled by the car while the brakes are applied is

d_3 = 247 m -216.2 m=30.8 m

We can find the acceleration of the car during this part by using the SUVAT equation:

v_f^2 - v_i^2 = 2ad_3

where

v_f = 0 is the final velocity (zero since the car comes to a stop)

v_i = 20.2 m/s is the velocity of the car at the moment the brakes are applied

a is the acceleration

d_3 = 30.8 m

Solving for a, we find

a=\frac{v_f^2 -v_i^2 }{2d}=\frac{0-(20.2)^2}{2(30.8)}=-6.62 m/s^2

5 0
3 years ago
Other questions:
  • mass and weight are similar, but not the same thing. In which of the following examples would the objects weight change, but mas
    8·1 answer
  • A capacitor has plates of area 8.25 * 10 ^ - 5 m ^ 2 . To create a capacitance of 3.35*10^ -10 F , how far apart should the plat
    8·1 answer
  • You observe three carts moving to the right. cart a moves to the right at nearly constant speed. cart b moves to the right, grad
    7·1 answer
  • Calculate the wavelength λ and the frequency f of the photons that have an energy of E photon = 2.32 × 10 − 19 J. Ephoton=2.32×1
    14·1 answer
  • Technician A says that low pressure smoke installed in the fuel system can be used to check for leaks. Technicians B says that n
    15·1 answer
  • The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
    8·1 answer
  • 14 What is the weight of each of
    5·1 answer
  • Examples for each different examples force affects​
    10·1 answer
  • I need help do in a hour
    6·1 answer
  • Now explore friction force. Set the piece of plastic or wood on the table and push it steadily across the tabletop using your fi
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!