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Radda [10]
4 years ago
12

Mercury thermometers work because mercury ____ when it is warmed

Chemistry
1 answer:
kap26 [50]4 years ago
5 0
Mercury expands when it is heated. This process is called thermal expansion.
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Which of the following contains the smallest amount of matter
ioda
The answer is 25g. Let me know if you need an explanation
5 0
3 years ago
What is the difference between a diatomic molecue and a normal molecue
Mekhanik [1.2K]

If a diatomic molecule consists of two atoms of the same element, such as hydrogen (H2) or oxygen (O2), then it is said to be homonuclear. Otherwise, if a diatomic molecule consists of two different atoms, such as carbon monoxide (CO) or nitric oxide (NO), the molecule is said to be heteronuclear.

7 0
4 years ago
Hello there!
Alenkasestr [34]

Answer:

Mass of carbon dioxide = 7.48 g

Explanation:

Given data:

Mass of lithium carbonate = 12.5 g

Mass of carbon dioxide produced = ?

Solution:

Chemical equation:

Li₂CO₃  →  Li₂O + CO₂

Number of moles of Li₂CO₃:

Number of moles = mass/ molar mass

Number of moles = 12.5 g /73.89 g/mol

Number of moles = 0.17 mol

Now we will compare the moles of Li₂CO₃  with CO₂.

                   Li₂CO₃        :            CO₂

                     1                :               1

                  0.17             :             0.17

Mass of carbon dioxide:

Mass of carbon dioxide = number of moles × molar mass

Mass of carbon dioxide =   0.17 mol ×  44 g/mol

Mass of carbon dioxide = 7.48 g

4 0
3 years ago
What do the COEFFICIENTS in a chemical reaction represent?
Alborosie

Answer:

the sum of the atoms or ions in the compound

5 0
4 years ago
A sample of gas is held at 1000C at a volume of 20 L. If the volume is increased to 40 L, what is the new temperature of the gas
kaheart [24]

Answer:

The new temperature will be 2546 K or 2273 °C

Explanation:

Step 1: Data given

The initial temperature = 1000 °C =1273 K

The volume = 20L

The volume increases to 40 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 20L

⇒with T1 = the initial temperature = 1273 K

⇒with V2 = the increased volume = 40L

⇒with T2 = the new temperature = TO BE DETERMINED

20L/ 1273 K = 40L / T2

T2 = 40L / (20L/1273K)

T2 = 2546 K

The new temperature will be 2546 K

This is 2546-273 = 2273 °C

Since the volume is doubled, the temperature is doubled as well

8 0
3 years ago
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