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3 years ago

Are the particles in a solid motionless ?

1 answer:

7 0

No. Just because it is a solid doesn't mean it has no motion. The only way for particles to be motionless is for the solid to reach Absolute Zero (0* K). Which is impossible. So, therefore, it is impossible for particles in a solid to be motionless.

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If an electron is transferred from A to B in the chemical reaction shown here,

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if an electron is passed from Ago B then Slides electrons which makes it electropositive i.e A+

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3 years ago

Calculate the formula mass of Cuso4

Mandarinka [93]

Answer: 159.61 g/mol

Copper has a molar mass of 63.546 g/mol

Sulfur has a molar mass of 32.065 g/mol

Oxygen has a molar mass of 16 g/mol, multiply by 4 and it is 16 x 4 = 64 g/mol

63.546 + 32.065 + 64 = 159.61 g/mol

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3 years ago

Write out the balanced equation for the reaction that occurs when Mg and Fe(NO3)3 react together. You do not need to make your s

Komok [63]

Balanced chemical reaction: 3Mg + 2Fe(NO₃)₃ → 2Fe + 3Mg(NO₃)₂.

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The reactivity series<span> is a series of metals from highest to lowest reactivity. Metal higher in the reactivity series will displace another, magnesium is more reactive than iron.</span>

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3 years ago

Provide the names of 2 stars that are dimmer than the Sun and provide their absolute magnitude and type of star.

Softa [21]

Explanation:

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5 0

3 years ago

Read 2 more answers

How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?

gulaghasi [49]

Answer:

$\boxed {\boxed {\sf 10.2 \ kJ}}$

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

$q=mc \Delta T$

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

m= 225 g
c= 0.897 J/g° C
ΔT= 50.5 °C

$q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

Multiply the first two numbers. The units of grams cancel.

$q= (225 \ g * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)$

$q= (225 * 0.897 \ J / \textdegree C)(50.5 \textdegree C)$

$q= (201.825\ J / \textdegree C)(50.5 \textdegree C)$

Multiply again. This time, the units of degrees Celsius cancel.

$q= 201.825 \ J * 50.5$

$q= 10192.1625 \ J$

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

$\frac { 1 \ kJ}{ 1000 \ J}$

Multiply by the answer we found in Joules.

$10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}$

$10192.1625 * \frac{ 1 \ kJ}{ 1000 }$

$\frac {10192. 1625}{1000} \ kJ$

$10.1921625 \ kJ$

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

$10.2 \ kJ$

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0

3 years ago