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Bezzdna [24]
3 years ago
10

Which particles make up the nucleus of an atom?

Chemistry
2 answers:
sveta [45]3 years ago
8 0
Protons and neutrons<span />
Anuta_ua [19.1K]3 years ago
3 0
Protons and neutrons
electrons just revolve around it
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Bari oxit có công thức hóa hóa học là BaO . vậy bari oxit được tạo nên từ các nguyên tố
Softa [21]

Answer:

Bari oxit được tạo nên từ các nguyên tố Bari (Ba) và Oxi (O)

Explanation:

6 0
2 years ago
How many moles of Ca atoms are in 1 mol of CaSO4?
Galina-37 [17]

Answer:

6.02 x 10²³ atoms

Explanation:

Given parameters:

Number of moles CaSO₄ = 1 mole

Unknown:

Number of Ca atoms in the given compound = ?

Solution:

The given compound is:

          CaSO₄

     1 mole of CaSO₄  is made up of 1 mole of Ca atoms

Now;

   1 mole of any substance contains 6.02 x 10²³ atoms

  1 mole of Ca atoms will also contain 6.02 x 10²³ atoms

7 0
3 years ago
What are miscible liquids?
sasho [114]

Answer:

ones that can be mixed together

Explanation:

like water or ethanol

7 0
3 years ago
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Methane is a type of coal.<br> true or false
Karolina [17]
The answer is false methane is emitted from coal
3 0
3 years ago
112 g of aluminum carbide react with 174 g water to produce methane and aluminum hydroxide in the reaction shown below.
dolphi86 [110]

<u>Answer:</u> 4.999 moles of excess reactant will be left over.

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       .....(1)

Given mass of aluminium carbide = 112 g

Molar mass of aluminium carbide = 143.96 g/mol

Putting values in equation 1:

\text{Moles of aluminium carbide}=\frac{112g}{143.96g/mol}=0.778mol

For the given chemical reaction:

2Al_4C_3(s)+12H_2O(l)\rightarrow 3CH_4(g)+4Al(OH)_3(s)

By the stoichiometry of the reaction:

2 moles of aluminium carbide reacts with 12 moles of water

So, 0.778 moles of aluminium carbide will react with = \frac{12}{2}\times 0.778=4.668 mol of water

Given mass of water = 174 g

Molar mass of water = 18 g/mol

Putting values in equation 1:

\text{Moles of water}=\frac{174g}{18g/mol}=9.667mol

Moles of excess reactant (water) left = 9.667 - 4.668 = 4.999 moles

Hence, 4.999 moles of excess reactant will be left over.

8 0
2 years ago
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