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Anastaziya [24]
3 years ago
7

How many joules are needed to change the temperate of 22g of water from 18°C to 33°C?

Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
5 0

Answer:

Q =  1379.4 J

Explanation:

Given data:

Mass of water = 22  g

Initial temperature = 18°C

Final temperature = 33°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 J/g. °C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 33°C - 18 °C

ΔT =  15°C

Q = 522 g ×4.18 J/g.°C× 15°C

Q =  1379.4 J

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MariettaO [177]

Answer:

The Bohr model show the protons in a carbon atom using the model;  "Electrons move in fixed orbits around a nucleus of protons and neutrons."

According to Bohr's model of the atoms, the Rutherford model is basically correct. This implies that Bohr model accepts the idea of a nucleus containing nucleons(protons and neutrons).

In addition, the model also postulates that electrons are found in fixed orbits. These fixed orbits are called energy levels or shells.

A graphic description of this is shown in the image attached to this answer.

Learn more: brainly.com/question/3964366

Explanation:

6 0
2 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

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4 0
3 years ago
Pls help asap!!!!
AVprozaik [17]

Answer: The pressure of the He is 2.97 atm

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

p_{total}=p_{N_2}+p_{O_2}+p_{He}.

Given : p_{total} =total pressure of gases = 6.50 atm

p_{N_2} = partial pressure of Nitrogen = 1.23 atm

p_{O_2} = partial pressure of oxygen = 2.3 atm

p_{He} = partial pressure of Helium = ?

putting in the values we get:

6.50atm=1.23atm+2.3atm+p_{He}  

p_{He}=2.97atm

The pressure of the He is 2.97 atm

3 0
3 years ago
Between which two elements is the difference in metallic character the greatest?
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Rb and O is your answer

6 0
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Rutherfordium has atomic number of 84 how many protons and electrons does it have
valentina_108 [34]

84   protons and 84 electrons.

5 0
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