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GenaCL600 [577]
3 years ago
11

If a 30 gram sample of Copper shots (specific heat = 0.385 J/gºC) changed from 27ºC to 90ºC, how much heat was involved?

Chemistry
2 answers:
Readme [11.4K]3 years ago
4 0

Answer:

A = 728J

Explanation:

Mass of substance = 30g

Specific capacity of copper = 0.385J/g°C

Initial temperature (T1) = 27°C

Final Temperature (T2) = 90°C

Heat energy (Q) = ?

Heat Energy (Q) = Mc∇T

Q = heat energy

M = mass of the substance

C = specific heat capacity of the substance

∇T = change in temperature of the substance = T2 - T1

Q = mc∇T

Q = mc×(T2 - T1)

Q = 30 × 0.385 × (90 - 27)

Q = 11.55 × 63

Q = 727.65J

The heat energy required to raise 30g of copper sample from 27°C to 90° is 728J

Natasha_Volkova [10]3 years ago
3 0

Answer:

728J

Explanation:

The following data were obtained from the question:

Mass (M) = 30g.

specific heat capacity (C) = 0.385 J/gºC.

Initial temperature (T1) = 27ºC

Final temperature (T2) = 90ºC

Change in temperature (ΔT) = T2 – T1 = 90°C – 27 = 63°C

Heat (Q) =...?

The heat involved can be obtained as illustrated below:

Q = MCΔT

Q = 30 x 0.385 x 63

Q = 728J

Therefore the heat involved is 728J

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Balance the following chemical equation:<br> C2H60+02 → CO2 + H20
pantera1 [17]

Answer:

Explain: In order to balance the chemical equation, you need to make sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. In order make both sides equal, you will need to multiply the number of atoms in each element until both sides are equal.

3 0
3 years ago
Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the
GarryVolchara [31]

Answer : The final temperature of the metal block is, 25^oC

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 aluminum = 55 g

m_2 = mass of water = 0.48 g

T_{final} = final temperature = ?

T_1 = temperature of aluminum = 25^oC

T_2 = temperature of water = 25^oC

c_1 = specific heat of aluminum = 0.900J/g^oC

c_2 = specific heat of water= 4.184J/g^oC

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

55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]

T_{final}=25^oC

Thus, the final temperature of the metal block is, 25^oC

6 0
3 years ago
The element californium (cf) sells for $1000 per µg. assuming 6.02 x 1023 atoms of cf have a mass of 251 grams, how many atoms o
saveliy_v [14]

Answer:- 2.40*10^1^0atoms

Solution:- It is a simple unit conversion problem. We could solve this using dimensional analysis.

We know that, 1 US dollar = 100 cents

1 cent  = 1 US penny

So, 1 US dollar = 100 US pennies

1g=10^6\mu g

Let's make the set up starting with 1 penny as:

1penny(\frac{$1}{100pennies})(\frac{1\mu g}{$1000})(\frac{1g}{10^6\mu g})(\frac{6.02*10^2^3atoms}{251g})

= 2.40*10^1^0atoms

Therefore, we can bye 2.40*10^1^0atoms of Cf in one US penny.

3 0
3 years ago
Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation: 4NH3 + 5O2 → 4NO + 6H2O Wh
tamaranim1 [39]

Answer:

16.16g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

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Mass of O2 from the balanced equation = 5 x 32 = 160g.

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2

Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.

Therefore, 16.16g of O2 is needed for the reaction.

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liubo4ka [24]

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Explanation:

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