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Natalija [7]
2 years ago
6

Two 10g blocks, one of copper and one of iron, were heated from 300 K to 400K (a temperature difference of 100 K).

Chemistry
1 answer:
12345 [234]2 years ago
6 0

1) 385 J

2) 450 J

Explanation:

1)

The amount of energy that must be absorbed by a certain substance in order to increase its temperature by \Delta T is given by the equation:

Q=mC\Delta T

where

m is the mass of the substance

C is its specific heat capacity

\Delta T is the increase in temperature of the substance

For the block of copper in this problem, we have:

m = 10 g is the mass

C=0.385 J/gK is the specific heat capacity of copper

\Delta T=400-300 = 100 K is the change in temperature

So, the energy absorbed by the block of copper is

Q=(10)(0.385)(100)=385 J

2)

Similarly for the block of iron, the energy absorbed by the iron is given by

Q=mC\Delta T

where

m is the mass of the block of iron

C is its specific heat capacity of iron

\Delta T is the increase in temperature of the block

Here we have:

m = 10 g is the mass of the block

C=0.450 J/gK is the specific heat capacity of iron

\Delta T=400-300 = 100 K is the change in temperature

So, the energy absorbed by the block of iron is

Q=(10)(0.450)(100)=450 J

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Answer:

Stage 1: 1 days.

Stage 2: 2-3 days.

Stage 3: 4-5 days.

Stage 4: 6 days.

Stage 5 (a-c): 7-12 days.

Stage 6: c. 17 days.

Stage 7: c. 19 days.

Stage 8: c. 23 days.

7 0
2 years ago
It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta
Zigmanuir [339]

Answer:

\lambda=241.9\ nm

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that  

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,  

1 mole = 6.023\times 10^{23}\ electrons

So,  

6.023\times 10^{23} electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of \frac {495.0}{6.023\times 10^{23}}\ kJ of energy.

Energy required = 82.18\times 10^{-23}\ kJ

Also,  

1 kJ = 1000 J

So,  

Energy required = 82.18\times 10^{-20}\ J

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}

\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}

\lambda=\frac{19.878}{10^6\times \:82.18}

\lambda=2.4188\times 10^{-7}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=241.9\ nm

6 0
2 years ago
Which of the following is true of group 6A?
USPshnik [31]

Answer:

The answer to your question is: Includes sulfur and gain two electrons

Explanation:

Includes Chlorine  This option is wrong, Chlorine belongs to group VII.

Includes Sulfur  This option is true, Group VI includes Oxygen, Sulfur, Selenium, Tellurium.

Gain 2 electrons . This option is true, Elements in group VI have six valence electrons so they gain to electrons to become estable.

Tend to form +2 ions  This option is wrong, this elements form -2 ions

Have 5 valence electrons This option is wrong, this elements have 6 valence electrons.

7 0
2 years ago
An inventor claims to have developed a new perfume that lasts a long time because it doesn't evaporate. Comment on this claim.
zaharov [31]

Answer:

C

Explanation:

Perfume needs to evaporate in order to smell.  If this perfume didn't evaporate, it would stay as a liquid and never smell.  

It wouldn't be D, as no toxic perfumes is sold.

It's not A because perfume doesn't have to be pressurized in order to not evaporate.

It's not B, as it is a hasty conclusion to the claim.  Plus, if the perfume did have an odor, even while not evaporating, the sales would be low as the product is that good.  

3 0
2 years ago
How many grams are there in 1.1 x 1027 molecules of water (H2O)?
Nezavi [6.7K]

Answer:

C) 3.3 x 104 grams

Explanation:

1 mole of water contains 6.02 × 10^23 atoms

1.1 × 10^27 atoms will contain;

1.1 × 10^27 ÷ 6.02 × 10^23

= 0.1827 × 10^( 27 - 23)

= 0.1827 × 10^(4)

= 1.827 × 10³ moles of water.

To convert mole to mass in grams, we use the formula;

mole (n) = mass (m) ÷ molar mass (MM)

Molar mas of water (H2O) = 1(2) of H + 16 of O = 18g/mol

mole = mass/molar mass

1.827 × 10³ = mass / 18

mass = 1.827 × 10³ × 18

mass = 32.886 × 10³

mass = 3.286 × 10⁴

mass = 3.3 × 10⁴ grams

6 0
2 years ago
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