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2 years ago

What is the concentration in %m/v of a 0.617 M aqueous solution of methanol (MM = 32.04 g/mol)?

Chemistry

1 answer:

svetoff [14.1K]2 years ago

6 0

Answer:

The correct answer is 1.977 % m/v ≅ 2% m/v

Explanation:

We have:

0.617 M = 0.617 moles methanol/ 1 L solution

We need:

%m/v= grams of methanol/100 mL solution

So, first we convert the moles of methanol to grams by using the MM (32.04 g/mol). Then, we multiply by 0,1 to convert the volume in liters to 100 mL by using the ratio: 100 mL= 0.1 L:

0.617 mol / 1 L x 32.04 g/mol 0.1 L/100 mL= 1.977 g/100 mL= %m/v

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A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a SoaveRedlich-Kwong gas and with

Setler79 [48]

Explanation:

The given data is as follows.

Moles of propylene = 100 moles, $T_{i}$ = 300 K

$T_{f}$ = 800 K, $V_{i} = 2 m^{3}$

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Now, we assume the following assumptions:

Since, it is a compression process therefore, work will be done on the system. And, work done will be equal to the heat energy liberating without any friction.

W = $mC_{p} \Delta T$

$100 moles \times 100 J/mol K (800 - 300) K$

= $5 \times 10^{6} J$

= 5 MJ

Thus, we can conclude that a minimum of 5 MJ work is required without any friction.

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2 years ago

A scientist is wondering why a certain region in the ocean doesn't have maximum phytoplankton growth, despite having plenty of n

Advocard [28]

Answer:

C The water had adequate nitrogen and phosphorus, so it is likely iron limited.

Explanation:

Phytoplankton are single- cell organisms that live in oceans.

They require nitrogen, phosphorus and trace amount of iron to survive.

From the scientist's results after testing the water for nitrogen and phosphorus,there are reasonable amount of these elements.

Therefore insufficient iron in the water is the reason why he could find plenty phytoplankton in the ocean.

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2 years ago

The laser in a 3D printer uses light with a frequency of 3.844 × 1014 Hz. Calculate the wavelength of this light in nm. Show all

Natali5045456 [20]

Answer:

The wavelength of this light is 780.4 nm.

Explanation:

Given that,

The frequency of laser light, $f=3.844 \times 10^{14}\ Hz$

We need to find the wavelength of this light.

We know that,

$c=f\lambda$

Where

$\lambda$ is the wavelength of light

So,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{3.844 \times 10^{14}}\\\\=7.804\times 10^{-7}\ m\\\\=780.4\ nm$

So, the wavelength of this light is 780.4 nm.

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1 year ago

What is the hybridization of the central atom in each of the following? 1. Beryllium chloride 2. Nitrogen dioxide 3. Carbon tetr

Lina20 [59]

Answer :

(1) The hybridization of central atom beryllium in $BeCl_2$ is, $sp$

(2) The hybridization of central atom nitrogen in $NO_2$ is, $sp^2$

(3) The hybridization of central atom carbon in $CCl_4$ is, $sp^3$

(4) The hybridization of central atom xenon in $XeF_4$ is, $sp^3d^2$

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

Now we have to determine the hybridization of the following molecules.

(1) The given molecule is, $BeCl_2$

$\text{Number of electrons}=\frac{1}{2}\times [2+2]=2$

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

(2) The given molecule is, $NO_2$

$\text{Number of electrons}=\frac{1}{2}\times [4]=2$

If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, $sp^2$ .

In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, $sp^2$ .

(3) The given molecule is, $CCl_4$

$\text{Number of electrons}=\frac{1}{2}\times [4+4]=4$

The number of electron pair are 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

(4) The given molecule is, $XeF_4$

$\text{Number of electrons}=\frac{1}{2}\times [8+4]=6$

Bond pair electrons = 4

Lone pair electrons = 6 - 4 = 2

The number of electrons are 6 that means the hybridization will be $sp^3d^2$ and the electronic geometry of the molecule will be octahedral.

But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.

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1 year ago

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masya89 [10]

Answer : The number placed in front of $CuSO_4$ should be, three (3).

Explanation :

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This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of chloride and sulfate ion are not balanced.

In order to balanced the chemical reaction, the coefficient 3 is put before the $CuSO_4$ , the coefficient 2 is put before the $AlCl_3$ and the coefficient 3 is put before the $CaCl_2$ .

Thus, the balanced chemical reaction will be,

$3CaSO_4+2AlCl_3\rightarrow 3CaCl_2+Al_2(SO_4)_3$

Therefore, the number placed in front of $CuSO_4$ should be, three (3).

5 0

2 years ago