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3 years ago

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Determine the electron geometry (eg) and molecular geometry (mg) of

1" title="CH_3^+" alt="CH_3^+" align="absmiddle" class="latex-formula">.
a) eg = bent, mg = bent

b) eg = tetrahedral, mg = trigonal planar

c) eg = trigonal pyramidal, mg = trigonal pyramidal

d) eg = tetrahedral, mg = tetrahedral

e) eg = trigonal planar, mg = trigonal planar

1 answer:

Yuri [45]3 years ago

7 0

Answer : The correct option is, (e) eg = trigonal planar, mg = trigonal planar

Explanation :

Formula used :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

The given molecule is, $CH_3^+$

$\text{Number of electron pair}=\frac{1}{2}\times [4+3-1]=3$

That means,

Bond pair = 3

Lone pair = 0

The number of electron pair are 3 that means the hybridization will be $sp^2$ and the electronic geometry of the molecule will be trigonal planar.

Hence, the electron geometry (eg) and molecular geometry (mg) of $CH_3^+$ is, trigonal planar and trigonal planar respectively.

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What was alchemy?

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Answer:

D

Explanation:

its a very old study and philosophy of how to change basic substances such as metals into other substances

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A solution of phosphoric acid was made by dissolving 10.0 g of H3PO4 in 100.0 mL of water. The resulting volume was 113 mL. Calc

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Explanation:

Mass of solute = 10.0 g

mass of solvent(water) = m

Volume of solvent( water) = v = 100.0 mL

Density of water= d = $1 g/cm^3=1 g/mL$

$1 mL= 1 cm^3$

$m=d\times v=1.0 g/mL\times 100.0 = 100.0 g$

Mass of solution(M) = Mass of solute + mass of solvent

M = 10.0 g + 100.0 g = 110.0 g

Volume of the solution = V = 113 mL

Density of the solution = D

$D=\frac{M}{V}=\frac{110.0 g}{113 mL}=0.9734 g/mL$

The density of the solution is 0.9734 g/ml.

Moles of phosphoric acid = $n_1=\frac{10.0 g}{98 g/mol}=0.1020 mol$

Moles of water = $n_2=\frac{100.0 g}{18g/mol}=5.556 mol$

Mole fraction of phosphoric acid = $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1020 mol}{0.1020 mol+5.556 mol}$

$\chi_1=0.01803$

Mole fraction of water = $\chi_2$

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{5.556 mol}{0.1020 mol+5.556 mol}$

$\chi_2=0.9820$

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

Moles of phosphoric acid = 0.1020 mol

Volume of the solution = V = 113 mL = 0.113 L ( 1 mL = 0.001 L)

Molarity of the solution :

$=\frac{0.1020 mol}{0.113 L}=0.903 M$

$[Molality]=\frac{\text{Moles of solute}}{\text{Mass of solvent(kg)}}$

Moles of phosphoric acid = 0.1020 mol

Mass of solvent(water) = m =100.0 g = 0.100 kg ( 1 g = 0.001 kg)

Molality of the solution :

$=\frac{0.1020 mol}{0.100 kg}=1.02 mol/kg$

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3 years ago

Which statement about subatomic particles is true? (2 points) Protons are the only subatomic particles to have charge. Electrons

Alona [7]

Protons are not the only subatomic particle that has a charge, electrons do to and they are negatively charged. Neutrons are situated in the nucleus of the atom, which means that they do not orbit it. Subatomic particles have different masses.

You are then left with Electrons are the subatomic particles with the smallest mass as your answer.

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3 years ago

If the point of the nail can be approximated as a circle with a radius 2.00×10^-3m What is the pressure in MPa exerted on the wa

Natasha_Volkova [10]

Answer:

8.28 MPa

Explanation:

From the question given above, the following data were obtained:

Radius (r) = 2×10¯³ m

Force applied (F) = 104 N

Pressure (P) =?

Next, we shall determine the area of the nail (i.e circle). This can be obtained as follow:

Radius (r) = 2×10¯³ m

Area (A) of circle =?

Pi (π) = 3.14

A = πr²

A = 3.14 × (2×10¯³)²

A = 3.14 × 4×10¯⁶

A = 1.256×10¯⁵ m²

Next, we shall determine the pressure. This can be obtained as follow:

Force applied (F) = 104 N

Area (A) = 1.256×10¯⁵ m²

Pressure (P) =?

P = F / A

P = 104 / 1.256×10¯⁵

P = 8280254.78 Nm¯²

Finally, we shall convert 8280254.78 Nm¯² to MPa. This can be obtained as follow:

1 Nm¯² = 1×10¯⁶ MPa

Therefore,

8280254.78 Nm¯² = 8280254.78 Nm¯² × 1×10¯⁶ MPa / 1 Nm¯²

8280254.78 Nm¯² = 8.28 MPa

Thus, the pressure exerted on the wall is 8.28 MPa

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3 years ago

When 45 g of an alloy, at 25oC, are dropped into 100.0 g of water, the alloy absorbs 956 J of heat. If the final temperature of

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Answer:

The answer to the question is

The specific heat capacity of the alloy = 1.77 J/(g·°C)

Explanation:

To solve this, we list out the given variables thus

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Final temperature of the alloy = 37 °C

Heat absorbed by the alloy = 956 J

Thus we have

ΔH = m·c·(T₂ - T₁) where ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C and m = mass of the alloy = 45 g

∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c or

c = 956 J/(540 g·°C) = 1.77 J/(g·°C)

The specific heat capacity of the alloy is 1.77 J/(g·°C)

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3 years ago