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aleksandrvk [35]
3 years ago
9

What volume units are greater than a liter

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
7 0
A gallon is greater than an liter hope this helps lol new to this
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How are temperature and thermal energy related
netineya [11]

HEAT IS RELATED TO THERMAL ENERGY AS THE PROCESS OF TRANSFERRING THERMAL ENERGY FROM ONE OBJECT OR SUBSTANCE TO ANOTHER.

HOPE THIS HELPS.

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3 years ago
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What unit is mass generally measured in
Veronika [31]

Answer:

kilograms

Explanation:

hope this helps, pls mark brainliest :D

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3 years ago
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start out as one kind of rock, but large amounts of pressure and heat change them into a different kind.
Eddi Din [679]

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Metamorphic

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6 0
3 years ago
How do you determine if and ion has a<br> noble-gasconfiguration?
Soloha48 [4]

Explanation:

Noble gas configuration is defined as the configuration which contains completely filled orbitals.

For example, oxygen atom when gain two electrons then it forms oxygen ion (O^{2-}).

Atomic number of oxygen atom is 8 and so, its number of electrons will also be 8. But when it gain two electrons then it has total 10 electrons. Hence, electronic configuration of O^{2-} is as follows.

          1s^{2}2s^{2}2p^{6}

Since, there are completely filled orbitals in an O^{2-} ions. Therefore, it means this ion has a  noble-gas configuration.

Thus, we can conclude that any specie which shows completely filled orbitals will have noble-gas configuration.

3 0
4 years ago
1. When 50.0 mL of water at 80.0°C was mixed with 50.0 mL of water in a calorimeter at
ddd [48]

Answer:

The heat capacity of the calorimeter is 5.11 J/g°C

Explanation:

Step 1: Given data

50.0 mL of water with temperature of 80.0 °C

Specific heat capacity of water = 4.184 J/g°C

Consider the density of water = 1g/mL

50.0 mL of water in a calorimeter at 20.0 °C

Final temperature = 47.0 °C

Step 2: Calculate specific heat capacity of the water in calorimeter

Q = Q(cal) + Q(water)

Q(cal) = mass * C(cal) * ΔT

Qwater = mass * Cwater * ΔT

Qcal = -Qwater

mass(cal) * C(cal) * ΔT(cal) =  mass(water) * C(water) * ΔT(water)

50 grams * C(cal) * (47.0 - 20.0) =- 50grams * 4.184 J/g°C * (47-80)

1350 * C(cal) = 6903.6

C(cal) = 5.11 J/g°C

The heat capacity of the calorimeter is 5.11 J/g°C

8 0
3 years ago
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