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Goshia [24]
3 years ago
14

Calculate the amount of heat gained when one 250 gram bottle is heated from 25oC to 30oC. The specific heat of water is 4.18 J/g

oC
Chemistry
2 answers:
Fed [463]3 years ago
7 0

Answer:

5230J

Explanation:

Mass (m) = 250g

Initial temperature (T1) = 25°C

Final temperature (T2) = 30°C

Specific heat capacity (c) = 4.184J/g°C

Heat energy (Q) = ?

Heat energy (Q) = Mc∇T

Q = heat energy

M = mass of the substance

C = specific heat capacity

∇T = change in temperature = T2 - T1

Q = 250 × 4.184 × (30 - 25)

Q = 1046 ×5

Q = 5230J

The heat energy required to raise the temperature of 250g of water from 25°C to 30°C is 5230J

Mariana [72]3 years ago
4 0

Answer:

The amount of heat gained is 5225 J

Explanation:

The specific heat of a substance is the quantity or amount of heat required to change the temperature of a unit mass of the substance by 1^{0}C.

Thus,

         Q = mcΔθ

where: Q is the quantity of heat required, m is the mass of the substance, c is the specific heat capacity of the substance and Δθ is the change in temperature of the substance.

From the question, given that: m = 250 g, initial temperature = 25^{0}C, final temperature = 30^{0}C and specific heat of water = 4.18 J/g^{0}C.

So that; the quantity of heat, Q, required is;

         Q = 250 × 4.18 × (30 - 25)

             = 250 × 4.18 × 5

            = 5225 Joules

⇒     Q = 5225 J

Therefore, the amount of heat gained is 5225 J.

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Explanation:

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Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76
aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.  

The no. of moles can be determined by using the formula,  

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

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The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.  

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,  

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,  

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.  

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