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taurus [48]
3 years ago
11

Determine the heat needed to warm 25.3 g of copper from 22 degrees celsius to 39 degrees celsius.

Chemistry
1 answer:
Serggg [28]3 years ago
5 0

Answer:

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

Explanation:

Q=mc\Delta T

Where:

Q = heat absorbed  or heat lost

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

We have mass of copper = m = 25.3 g

Specific heat of copper = c = 0.385 J/g°C

ΔT  = 39°C - 22°C = 17°C

Heat absorbed by the copper :

Q=25.3 g\times 0.385 J/g^oC\times 17^oC=165.59 J

The heat needed to warm 25.3 g of copper from 22°C to 39°C is 165.59 Joules.

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What is the atomic mass of a carbon isotope that has 6 protons and 7 neutrons?
DedPeter [7]

The atomic mass of a carbon isotope that has 6 protons and 7 neutrons is<u> </u><u>13</u>

Explanation:

Remember that whilst the atomic number represents the number of protons in an atom, the mass number represents the summation of protons and neutrons particles in the atomic nuclei. Therefore, in this case, the carbon will have a mass number of;

6 +  7 = 13

Isotopes of an element usually have the same atomic number but different mass numbers -because they have slightly different numbers of neutrons. An example is isotopes of Carbon; C-14 and C-12

7 0
3 years ago
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A single atom of platinum has a mass of 3.25 x 10-22g. What is the mass of 6.0 x 1023 platinum atoms?
boyakko [2]
Even though the numbers are in scientific notation, they are still just numbers. Essentially the question is as simple as "An apple weighs an ounce, how much do 35 apples weigh?". So all you have to do is multiply.

3.25 x 10^{-22}   x  6.0 x10^{23}  = 195

6 0
3 years ago
Even though lead is toxic, lead compounds were used in ancient times as white pigments in cosmetics. What is the percentage of l
Gekata [30.6K]

<u>Answer:</u> The mass percent of lead in lead (IV) carbonate is 63.32 %

<u>Explanation:</u>

The given chemical formula of lead (IV) carbonate is Pb(CO_3)_2

To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:

\text{Mass percent of lead}=\frac{\text{Mass of lead}}{\text{Mass of lead (IV) carbonate}}\times 100

Mass of lead = (1 × 207.2) = 207.2 g

Mass of lead (IV) carbonate = [(1 × 207.2) + (2 × 12) + (6 × 16)] = 327.2 g

Putting values in above equation, we get:

\text{Mass percent of lead}=\frac{207.2g}{327.2g}\times 0100=63.32\%

Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %

4 0
2 years ago
What is the approximate size of solute particles in a solution?
jeyben [28]

Answer:

The answer is B

Explanation:

8 0
3 years ago
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Please help with this
dlinn [17]

Answer:

A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.

Step by Step Explanation?

Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.

The formula is,

P₁ x V₁ = P₂ × V₂

Where,

P₁ is initial pressure = 1 atm

P2 is final pressure = ? (Not Known)

V₁ is initial volume = 10 L

V₂ is final volume = 15 L

Now put the values in the formula,

\begin{gathered}\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67\end{gathered]

Therefore, the answer is 0.67 atm.

5 0
2 years ago
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