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Dmitriy789 [7]
2 years ago
15

If 300. mL of water are poured into the measuring cup, the volume reading is 10.1 oz . This indicates that 300. mL and 10.1 oz a

re equivalent. How many milliliters are in a fluid ounce based on this data?
Physics
2 answers:
tigry1 [53]2 years ago
5 0

Answer:

Milliliters to Ounces Conversions

some results rounded

mL    - fl oz

200.00 6.7628

200.01 6.7631

200.02 6.7635

200.03 6.7638

200.04 6.7642

200.05 6.7645

200.06 6.7648

200.07 6.7652

200.08 6.7655

200.09 6.7658

200.10 6.7662

200.11 6.7665

200.12 6.7669

200.13 6.7672

200.14 6.7675

200.15 6.7679

200.16 6.7682

200.17 6.7686

200.18 6.7689

200.19 6.7692

200.20 6.7696

200.21 6.7699

200.22 6.7702

200.23 6.7706

200.24 6.7709

mL fl oz

200.25 6.7713

200.26 6.7716

200.27 6.7719

200.28 6.7723

200.29 6.7726

200.30 6.7729

200.31 6.7733

200.32 6.7736

200.33 6.7740

200.34 6.7743

200.35 6.7746

200.36 6.7750

200.37 6.7753

200.38 6.7757

200.39 6.7760

200.40 6.7763

200.41 6.7767

200.42 6.7770

200.43 6.7773

200.44 6.7777

200.45 6.7780

200.46 6.7784

200.47 6.7787

200.48 6.7790

200.49 6.7794

mL fl oz

200.50 6.7797

200.51 6.7800

200.52 6.7804

200.53 6.7807

200.54 6.7811

200.55 6.7814

200.56 6.7817

200.57 6.7821

200.58 6.7824

200.59 6.7828

200.60 6.7831

200.61 6.7834

200.62 6.7838

200.63 6.7841

200.64 6.7844

200.65 6.7848

200.66 6.7851

200.67 6.7855

200.68 6.7858

200.69 6.7861

200.70 6.7865

200.71 6.7868

200.72 6.7872

200.73 6.7875

200.74 6.7878

mL fl oz

200.75 6.7882

200.76 6.7885

200.77 6.7888

200.78 6.7892

200.79 6.7895

200.80 6.7899

200.81 6.7902

200.82 6.7905

200.83 6.7909

200.84 6.7912

200.85 6.7915

200.86 6.7919

200.87 6.7922

200.88 6.7926

200.89 6.7929

200.90 6.7932

200.91 6.7936

200.92 6.7939

200.93 6.7943

200.94 6.7946

200.95 6.7949

200.96 6.7953

200.97 6.7956

200.98 6.7959

200.99 6.7963

Explanation:

Levart [38]2 years ago
5 0

Answer : 29.7 mL are in a fluid ounce based on this data.

Explanation :

As we are given that 300 mL and 10.1 oz are equivalent. That means,

300 mL = 10.1 oz

or,

10.1 oz = 300 mL

Now we have to determine the volume of fluid in milliliters.

As, 10.1 oz of fluid = 300 mL

So, 1 oz of fluid = \frac{1oz}{10.1oz}\times 300mL

                         = 29.7 mL

Therefore, 29.7 mL are in a fluid ounce based on this data.

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Calculate the number of moles in each of the following masses: 0.039 g of palladium 0.0073 kg of tantalum
marysya [2.9K]

Answer:

<em>The number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>

Explanation:

Number of mole = reacting mass/molar mass

n = R.m/m.m......................... Equation 1

Where n = number of moles, R.m = reacting mass, m.m = molar mass.

For palladium,

R.m = 0.039 g and m.m = 106.42 g/mol

Substituting theses values into equation 1

n = 0.039/106.42

n = 0.00037 mole

For tantalum,

R.m = 0.0073 and m.m = 180.9 g/mol

Substituting these values into equation 1

n = 0.0073/180.9

n = 0.0000404 mole

<em>Therefore the number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>

3 0
2 years ago
If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat
Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

3 0
2 years ago
A metallic sphere of radius 2.0 cm is charged with +5.0-μC+5.0-μC charge, which spreads on the surface of the sphere uniformly.
sladkih [1.3K]

Answer:

Explanation:

Potential due to a charged metallic sphere having charge Q and radius r on its surface will be

v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre  potential is

v = k Q / R

a ) On the surface of the shell , potential due to positive charge is

V₁ = \frac{9\times10^9\times5\times10^{-6}}{6\times10^{-2}}

On the surface of the shell , potential due to negative  charge is

V₁ = \frac{- 9\times10^9\times5\times10^{-6}}{6\times10^{-2}}

Total potential will be zero . they will cancel each other.

b ) On the surface of the sphere potential

= \frac{9\times10^9\times5\times10^{-6}}{2\times10^{-2}}

= 22.5 x 10⁵ V

On the surface of the sphere potential due to outer shell

= \frac{9\times10^9\times5\times10^{-6}}{5\times10^{-2}}

= -9 x 10⁵

Total potential

=( 22.5 - 9 ) x 10⁵

= 13.5 x 10⁵ V

c ) In the space between the two , potential will depend upon the distance of the point from the common centre .

d ) Inside the sphere , potential will be same as that on the surface that is

13.5 x 10⁵ V.

e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.

5 0
2 years ago
A 1700kg rhino charges at a speed of 50.0km/h. what is the magnitude of the average force needed to bring the rhino to a stop in
stepladder [879]
Impulse equals Change in Momentum
F = average applied force = to be determined
Δt = time during which the force is applied = 0.50 s
m = mass = 1,700 kg
Δp = change in momentum = to be determined
Δv = change in velocity = to be determined
v1 = initial velocity = 50.0 km/h = 50,000 m/h = 13.9 m/s
v2 = final velocity = 0.00 km/h = 0.00 m/s

F∙Δt = Δp
F∙Δt = m∙Δv
F∙Δt = m∙(v2 - v1)

F = m∙(v2 - v1) / Δt
F = 1,700 kg∙(0.00 m/s - 13.9 m/s) / 0.50 s
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7 0
2 years ago
The trough of the sine curve used to represent a sound wave corresponds to
iren [92.7K]

Answer:

The correct answer is a rarefaction.

Explanation:

Sound waves are longitudinal waves that propagate in a medium, such as air. As the vibration continues, a series of successive condensations and rarefactions form and propagate from it. The pattern created in the air is something like a sinusoidal curve to represent a sound wave.

There are peaks in the sine wave at the points where the sound wave has condensations and valleys where it has rarefactions.

Have a nice day!

4 0
2 years ago
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