Two processes are described below:

Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w

just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0

3 years ago

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

3 years ago

Read 2 more answers

A gas occupies 25,3 mL at a pressure of 152 kPa. Find the volume if the pressure is

sleet_krkn [62]

Answer:

This is under gas laws. check it out

3 0

3 years ago

Which has more thermal energy, 500g of ice or 500g of steam? explain your answer.

RUDIKE [14]

Steam because steam comes from water which the energy makes it into steam

7 0

4 years ago

PLEASEEEE HELP MEEEEE

fiasKO [112]

Answer:

3.82 x 10²¹ molecules As₂O₃

Explanation:

To find the amount of molecules arsenic (III) oxide (As₂O₃), you need to (1) convert kg to lbs, then (2) convert g As₂O₃ to moles As₂O₃ (via molar mass), and then (3) convert moles to molecules (via Avogadro's number).

1 kilogram = 2.2 lb

Molar Mass (As₂O₃): 2(74.992 g/mol) + 3(15.998 g/mol)

Molar Mass (As₂O₃): 197.978 g/mol

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

0.0146 g As₂O₃ 1 kg 189 lb
------------------------ x --------------- x ------------------ x ................
1 kg 2.2 lb

1 mole 6.022 x 10²³ molecules
x ------------------ x --------------------------------------- = 3.82 x 10²¹ molecules As₂O₃
197.978 g 1 mole

6 0

2 years ago

Read 2 more answers