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skelet666 [1.2K]
3 years ago
15

How are scientific questions answered?

Chemistry
2 answers:
dem82 [27]3 years ago
7 0

Answer:

B .Through testing a theory about the physical world

Explanation:

BigorU [14]3 years ago
7 0

Answer:

Option B

Explanation:

Scientific question are answered through experimentation, through testing the theory about the physical world.

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In ionic bonding, during the transfer of electrons between two neutrally charged atoms, one electron moves from one atom to anot
forsale [732]

The given question is incomplete . The complete question is :

In ionic bonding, during the transfer of electrons between two neutrally charged atoms, one electron moves from one atom to another. What are the new relative charges between the two atoms?

a. The giving atom and receiving atom are both negatively charged

b. The giving atom is now positively charged and the receiving atom is now negatively charged.

c. The giving and receiving atom are both positively charged

d. The giving atom is now negatively charged and the receiving atom is now positively charged.

Answer:  The giving atom is now positively charged and the receiving atom is now negatively charged.

Explanation:

Ionic compounds are formed by transference of electrons between metals and non metals. The bond formed between a metal and a non-metal is always ionic in nature.

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Thus when  one electron moves from one atom to another, the metal acquires a positive charge and the non metal acquires negative charge.

For example:  NaCl is formed by transfer of one electron from sodium to chlorine , thus forming Na^+ and Cl^-

3 0
3 years ago
Draw the dot structure of CX4
hichkok12 [17]

Refer to the attachment

7 0
3 years ago
How many atoms are in a sample of 1.83 moles of potassium (K) atoms? Please explain the conversions to me. Thank you!
IRINA_888 [86]
1 mole K ------------- 6.02x10²³ atoms
1.83 moles K ------ ?? atoms

1.83 x (6.02x10²³) / 1 =

1.101x10²⁴ atoms of K

hope this helps!
6 0
3 years ago
The half life of plutonium-239 (239pu, pu-239) is 24,100 years. How much plutonium will remain after 1000 years if the initial a
Anestetic [448]

The mass of plutonium that will remain after 1000 years if the initial amount is 5 g when the half life of plutonium-239 (239pu, pu-239) is 24,100 years is 2.5 g

The equation is Mr=Mi(1/2)^n

where n is the number of half-lives

Mr is the mass remaining after n half lives

Mi is the initial mass of the sample

To find n, the number of half-lives, divide the total time 1000 by the time of the half-life(24,100)

n=1000/24100=0.0414

So Mr=5x(1/2)^1=2.5 g

The mass remaining is 2.5 g

  • The half life is the time in which the concentration of a substance decreases to half of the initial value.

Learn more about half life at:

brainly.com/question/24710827

#SPJ4

8 0
1 year ago
Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...
g100num [7]

Answer:

O₂; KCl; 33.3  

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

            2KCl  +  3O₂ ⟶ 2KClO₃

n/mol:  100.0   100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}

(ii) From O₂

\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

3 0
3 years ago
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