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3 years ago

7

HELP PLEASE

1 answer:

-BARSIC- [3]3 years ago

7 0

Angel ! You have a formula, and you have an example that's
completely worked out. The ONLY POSSIBLE reason that you
could still need help is that you're letting math scare you.

I'll do 'A' for you, 'B' most of the way, and get 'C' set up.
If THAT's not enough for you to run with and finish them all,
then you and I should both be embarrassed.

Write the formula on the wall:

   °F = (9/5) °C + 32°

A). Convert 35° C °F = (9/5)(35°) + 32°

(9/5)(35) = 63 °F = 63° + 32°

   °F = 95°
____________________________________

B). Convert 80°F to °C
       The formula: °F = (9/5) °C + 32°

°F = 80    80 = (9/5)°C + 32

Subtract 32 from each side: 48 = (9/5)°C

Multiply each side by 5 : 240 = (9) C

Now you take over:
_________________________________________

C). Convert 15°C to °F.
       The formula: °F = (9/5) °C + 32°

°C = 15 °F = (9/5) 15° + 32

(9/5) (15) = 27
   Go !    °F =

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A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

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3 years ago

Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the

Scrat [10]

Answer: $1.8\°$

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d=3.4(10)^{-5}m$ is the width of the slit

$\lambda=540 nm=540(10)^{-9}m$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the second-order diffraction angle is given when $n=2$ , hence equation (1) becomes:

$dsin\theta_{2}=2\lambda$ (2)

Now we have to find the value of $\theta_{2}$ :

$sin\theta_{2}=\frac{2\lambda}{d}$ (3)

Then:

$\theta_{2}=arcsin(\frac{2\lambda}{d})$ (4)

$\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})$ (5)

Finally:

$\theta_{2}=1.8\°$ (6)

5 0

3 years ago