Why is an energy source needed in order to have a working electric circuit

Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the

NikAS [45]

The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

Answer:

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

Explanation:

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is $ns^{1-2}$

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is $np^{1-6}$

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is $[(n-1)d^{1-10}ns^{0-2}]$

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is $[(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]$ . They are also known as lanthanide and actinide series.

For the given elements:

Option 1: Hf

Hafnium is the 72nd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^26s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 2: Am

Americium is the 95th element of the periodic table having electronic configuration of $[Rn]5f^{7}6d^07s^2$

As, the last electron is entering the (f subshell), it is a inner transition element.

Option 3: In

Indium is the 49th element of the periodic table having electronic configuration of $[Kr]5s^25p^1$

As, the last electron is entering the p subshell, it is a main group element.

Option 4: Ta

Tantalum is the 73rd element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^56s^2$

As, the last electron is entering the d subshell, it is a transition element.

Option 5: As

Arsenic is the 33rd element of the periodic table having electronic configuration of $[Ar]4s^24p^3$

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

Option 6: Se

Selenium is the 34th element of the periodic table having electronic configuration of $[Ar]4s^24p^4$

As, the last electron is entering the p subshell, it is a main group element.

Option 7: Rn

Radon is the 86th element of the periodic table having electronic configuration of $[Xe]4f^{14}5d^{10}6s^26p^6$

As, the last electron is entering the p subshell, it is a main group element.

5 0

3 years ago

EL hidroxido de sodio reacciona con el sulfato de hierro (II) para formar sulfato de sdio e hidroxido ferroso. Si se hace reacci

Oduvanchick [21]

Answer:

280.8 g

Explanation:

Definimos la reaccion:

2NaOH + FeSO₄ → Na₂SO₄ + Fe(OH)₂

Como tenemos la masa de NaOH, asumimos que el sulfato de hierro (II) es el reactivo en exceso.

Definimos masa de reactivo: 250 g . 1mol / 40g = 6.25 mol

2 moles de NaOH producen 1 mol de hidroxido ferroso

Entonces 6.25 moles producirán, la mitad (6.25 . 1) /2 = 3.125 moles

Convertimos los moles a masa:

3.125 mol . 89.85 g/mol = 280.8 g

7 0

3 years ago

How many moles are there in 24.00 g of NaCl

Elis [28]

Answer:

The answer to your question is 0.41 moles

Explanation:

Data

moles of NaCl = ?

mass of NaCl = 24 g

Process

To solve this problem just calculate the molar mass of NaCl, and remember that the molar mass of any substance equals to 1 mol.

1.- Calculate the molar mass

NaCl = 23 + 35.5 = 58.5 g

2.- Use proportions and cross multiplication

58.5 g of NaCl ------------------- 1 mol

24.0 g ------------------- x

x = (24 x 1) / 58.5

x = 0.41 moles

6 0

3 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago

Write the value represented by 4.5x10-6 g

monitta

the value represented by 4.5x10-6 g is 0.0000045 g it is a representation in standard notation

To express enormous figures which are in standard notation like 1,300,000 or exceedingly minuscule quantities like 0.0000000000045, scientists use scientific notation. Scientific notation, also referred to as exponential form, is one of the earliest mathematical techniques. It is well regarded by practitioners. People use scientific notation to handle situations where numbers are too large or too small to be calculated easily. Scientists, engineers, and mathematicians all use this technique. where as standard notation is way writing numbers in normal decimal form. The scientific method id most convenient denoting numbers as the numbers will be either too big or too small.

To learn more about standard notation:

brainly.com/question/10253124

#SPJ4

6 0

1 year ago