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lisov135 [29]
3 years ago
8

What is the mass of an element with 15 protons, 13 electrons, and 11 neutrons? *

Chemistry
1 answer:
RSB [31]3 years ago
6 0

Answer:

The mass of the element is 26.20 amu

Explanation:

In this question, we are asked to calculate the mass of an element with 15 protons, 13 electrons and 11 neutrons

To calculate the atomic mass of the element, we take into consideration the masses of the individual sub atomic particles

Electrons have 0 atomic mass unit(their masses are negligible) we have no business here, Protons have a mass of

1.00727647 amu, while the mass of neutron is 1.0086654 amu

The mass of 15 protons is thus 15 * 1.00727647 = 15.10914705 amu

The mass of 11 neutrons is 11 * 1.0086654 =

11.0953194 amu

Adding this together, we have ; 11.0953194 + 15.10914705 = approximately 26.20 amu

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Is sisig is homogeneous or heterogeneous​
givi [52]

Answer:

I think its homogeneous

Explanation:

because its all pig, just different parts of the pig

7 0
3 years ago
H3PO4 + Ca(OH)2 → Ca(H2PO4)2 + H2O
aniked [119]

Given question is incomplete. The complete question is as follows.

Balance the following equation:

H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + H_2O

Answer: The balanced chemical equation is as follows.

2H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + 2H_2O

Explanation:

When a chemical equation contains same number of atoms on both reactant and product side then this equation is known as balanced equation.

For example, H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + H_2O

Number of atoms on reactant side:

H = 5

P = 1

O = 6

Ca = 1

Number of atoms on product side:

H = 6

P = 2

O = 9

Ca = 1

In order to balance this equation, we will multiply H_3PO_4 by 2 on reactant side and we will multiply H_2O by 2 on product side. Hence, the balanced chemical equation is as follows.

2H_3PO_4 + Ca(OH)_2 \rightarrow Ca(H_2PO_4)_2 + 2H_2O

8 0
3 years ago
If you knew the number of valence electrons in a nonmetal atom how would you determine the valence of the element. (Ignore hydro
SIZIF [17.4K]

Answer:

The possible valances can be determined by electron configuration and electron negativity

Good Luck even though this was asked 2 weeks ago

Explanation:

All atoms strive for stability. The optima electron configuration is the electron configuration of the VIII A family or inert gases.

Look at the electron configuration of the nonmetal and how many more electrons the nonmetal needs to achieve the stable electron configuration of the inert gases. Non metals tend to be negative in nature and gain electrons. ( They are oxidizing agents)

For example Florine atomic number 9 needs one more electron to reach a valance number of 8 electrons to equal Neon atomic number 10. Hence Flowrine has a valance of -1

Oxygen atomic number 8 needs two more electrons to reach a valance number of 8 electrons to equal Neon atomic number 10. Hence Oxygen has a valance charge of -2.

Non metals with a low electron negativity will lose electrons when reacting with another non metal that has a higher electron negativity. When the non metal forms an ion it is necessary to look at the electron structure to determine how many electrons the element can lose to gain stability.

For example Chlorine which is normally -1 like Florine when it combines with oxygen can be +1, +3, + 5 or +7. It can lose its one unpaired electron, or combinations of the unpaired electron and sets of the three pairs of electrons.

6 0
2 years ago
Consider this equilibrium:
Ilia_Sergeevich [38]

Answer:

the answer is option E they are bronsted lowry acid

7 0
2 years ago
How many moles of CCI are there in 78.2 g of CCI.?
vaieri [72.5K]

Answer:

0.508 mole

Explanation:

NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.

The number of mole present in 78.2 g of CCl₄ can be obtained as follow:

Mass of CCl₄ = 78.2 g

Molar mass of CCl₄ = 12 + (35.5×4)

= 12 + 142

= 154 g/mol

Mole of CCl₄ =?

Mole = mass / molar mass

Mole of CCl₄ = 78.2 / 154

Mole of CCl₄ = 0.508 mole

Therefore, 0.508 mole is present in 78.2 g of CCl₄

6 0
2 years ago
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