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allochka39001 [22]
3 years ago
12

What does the space-filling model of a water molecule tell you about the relative size of the atoms?

Chemistry
1 answer:
faltersainse [42]3 years ago
8 0
Space -filling models are also known as Calotte models. These are 3 dimensional models that depict the spatial relationships between atoms.
The space-filling model of water molecule shows the 3 D structure of the water molecule. This model CLEARLY REVEALS THE OXYGEN ATOM WHICH IS LOCATED CENTRALLY WITH TWO HYDROGEN ATOMS IN THE ADJACENT SIDES. THE MODEL ALSO SUGGESTS HOW THE WATER MOLECULES ATTRACT EACH OTHER.
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Write the appropriate symbol for each of the following isotopes: (a) Z 11, A 23; (b) Z= 28, A= 64; (c) Z= 50, A =115; (d) Z= 20,
prohojiy [21]

Explanation:  

Z = atomic mass of the element and  , A = atomic mass of the element .

a) Z = 11, A =  23

Element = Sodium

  symbol: ²³₁₁Na  .

b) Z = 28, A =  64

Element = Nickel

  symbol: ⁶⁴₂₈Ni  .

c) Z = 50, A = 115

Element = tin

  symbol: ¹¹⁵₅₀Sn  .

d) Z = 20, A = 42

Element = Calcium

  symbol: ⁴²₂₀Ca .

6 0
3 years ago
Read 2 more answers
What is the easiest way to explain molecular compounds and covalent bonds also adding information about valance electrons?
wlad13 [49]

Answer:

Molecular compounds consist of two or more nonmetals. The nonmetals that make up a molecular compound are held together by covalent/molecular bonds. Covalent bonds is known as the "sharing" of valence electrons between two or more chemical species. Valence electrons are shared so that the atoms of the compound can become stable, much like how ionic bonds transfer valence electrons between atoms to achieve stability.

5 0
3 years ago
Consider the titration of sulfuric acid with sodium hydroxide. What volume (mL) of a 2.658M NaOH solution is required to fully t
KatRina [158]

The volume of base required to completely neutralize the acid is 3.2 mL of NaOH.

The equation of the reaction is;

2NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)

From the question;

Concentration of acid CA = 0.426M

Concentration of base  CB = 2.658M

Volume of acid VA = 10.00mL

Volume of base VB = ?

Number of moles of acid NA = 1

Number of moles of base NB = 2

Using the relation;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

VB = CAVANB/CBNA

VB =  0.426M × 10.00mL × 2/ 2.658M × 1

VB = 3.2 mL

Learn more: brainly.com/question/6111443

8 0
2 years ago
Determining the composition of stars is one way that light emission has been used to determine the chemical composition of an ob
Blizzard [7]

Answer:

Explanation:

These instrument works on the analysis of the emisson spectral of light received from the star in this way.

Think of a steel knife in your kitchen. Initially, it has this shiny silver colour that typifies it. When the knife is placed on a hot plate, it becomes hotter and begins to go red as the heating continues. If we stop the heating and pour cold water on it, the red dissapears and our knife is back to itself, although the silvery shine would be lost. This is simply how the atomic absorption spectroscopy works. When you see the hot knife you can say a couple of things about it. Different metals have their various melting point. We can compare the temperature at which our knife will melt with a standard melting point scale to know the type of metal it is made of.

In atomic absorption spectroscopy, an atom gains energy and it becomes excited. Every atom is known to have a peculair amount of absorbant energy that cause them to excite. The more the particles in the atom, the more the energy required. When we analyse the absorbent energy of the atom, it differs from other atoms and we truly identify such an atom even if we don't know it. Most times, the energy is given off as light.

3 0
3 years ago
SO
Burka [1]

Answer:

\large \boxed{\text{E) 721 K; B) 86.7 g}}

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL;  V₂ = 435 mL

T₁ = 355 K;     T₂ = ?

b) Calculation

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}

7 0
3 years ago
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