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Furkat [3]
3 years ago
14

A chunk of copper has a volume of 1.45 mL and a mass of 12.93 grams. What is the density of the chunk of copper?

Chemistry
2 answers:
babunello [35]3 years ago
8 0

Answer:

8.92 g/ml

Explanation:

Use the formula: density = mass/volume and substitute the values:

density = 12.93 ÷ 1.45 = 8.91724... ≈ 8.92 g/ml

The density of the chunk of copper is 8.92 g/ml

Hope this helps!

natta225 [31]3 years ago
7 0

Answer:

Density = 8.92 \ g/mL

Explanation:

The equation for density is:

Density = \frac{mass}{volume}

We plug in the given values:

Density = \frac{12.93 \ g}{1.45 \ mL}

Density = 8.92 \ g/mL

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Consider the dissolution of ammonium nitrate. 1.25 g of amminum nitrate is dissolved in enough water to make 25.0 mL of solution
Mrac [35]

Answer:

\Delta H=-0.02\ kJ

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

\Delta H=m\times C\times \Delta T

Where,

\Delta H  is the enthalpy change

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass of ammonium nitrate = 1.25 g

Specific heat = 4.18 J/g°C

\Delta T=21.9-25.8\ ^0C=-3.9\ ^0C

So,

\Delta H=-1.25\times 4.18\times 3.9\ J=-20.3775\ J

Negative sign signifies loss of heat.

Also, 1 J = 0.001 kJ

So,

\Delta H=-0.02\ kJ

5 0
3 years ago
Would you expect a butane lighter to work in winter when the temperature outdoors is 25 ∘F?
kogti [31]
No, I would not expect the lighter to work at 25 °F.

25 °F ≈ -4 °C

The boiling point of butane is -1 °C.

At -4 °C, the vapour pressure of butane is not great enough to push back against atmospheric pressure.

The lighter won't work.
7 0
3 years ago
Science hmk due tmr morning need help!! pls answer all 3 and number which one is which for example...
MaRussiya [10]
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2)A because potential energy is always at the top first
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8 0
3 years ago
A liquid takes 10.14 x 10^6 J of energy to boil 28.47 kg at 298 K. Using the latest heats of vaporization, what substance is thi
otez555 [7]

A liquid takes 10.14 x 10^6 J of energy to boil 28.47 kg at 298 K. Using the Latent heats of vaporization of some of the substances:

Acetone: 538,900 \frac{J}{kg}

Ammonia: 1,371,000 \frac{J}{kg}

Propane: 356,000 \frac{J}{kg}

Methane: 480,600 \frac{J}{kg}

Ethanol: 841,000 \frac{J}{kg}

Calculating the latent heat of vaporization of the given substance:

Heat required = 10.14 * 10^{6} J

Mass of the substance boiled = 28.47 kg

Latent heat of vaporization =\frac{10.14 * 10^{6} J}{28.47 kg}  = 3,56,000 \frac{J}{kg}

So the substance boiled is propane.


5 0
3 years ago
describe, in terms of electrons, what happens when a magnesium atom reacts with chlorine atoms to produce magnesium chloride?
BigorU [14]
The metal ions become positive ions and the non-metal atoms become negative ions. There is a strong electrostatic force of attraction between these oppositely charged ions called an ionic bond.
3 0
4 years ago
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