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egoroff_w [7]
3 years ago
5

calculate the charge on plates if the plate area is 100 cm2 , the gap between plates is 10 mm and the scale is reading 1.0 gram.

Chemistry
1 answer:
Law Incorporation [45]3 years ago
8 0

Answer: The charge on the plates are 88.4 picafarad

Explanation:The equation used in measuring charge in a plate is given as:

C=Q/V =E A/D

Where E= dielectric content

A= Area of plates

d= distance between plates

Using dielectric constant for Air=8.84×10-12F/m

A=100cm2=0.01m2

d=10mm=0.001m

C= 8.84×10-12×0.01/0.001

C= 88.4 picafarad

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Your answer would be 0.00285 moles.
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A rubber balloon was filled with helium at 25.0˚C and placed in a beaker of liquid nitrogen at -196.0˚C. The volume of the cold
Ksenya-84 [330]

Answer:

The volume of helium at 25.0 °C is 60.3 cm³.

Explanation:

In order to work with ideal gases we need to consider absolute temperatures (Kelvin). To convert Celsius to Kelvin we use the following expression:

K = °C + 273.15

The initial and final temperatures are:

T₁ = 25.0 + 273.15 = 298.2 K

T₂ = -196.0 + 273.15 = 77.2 K

The volume at 77.2 K is V₂ = 15.6 cm³. To calculate V₁ in isobaric conditions we can use Charle's Law.

\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}} \\V_{1}=\frac{V_{2}}{T_{2}} \times T_{1}=\frac{15.6cm^{3} }{77.2K} \times 298.2K=60.3cm^{3}

3 0
3 years ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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