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2 years ago

Why is it important to have regular supervision of the weight and measurements in the market

Chemistry

1 answer:

Sergio [31]2 years ago

4 0

Answer:

Supervision of weights and measures promotes accurate measurements of goods and services to ensure that everybody gets a fair trade in the marketplace. Not so coincidentally it also is a deterrent to ensure that traders are being honest in their trade practises.

Explanation:

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Oxide formation by heating element in oxygen atmosphere are best classified as which type of reaction?a. Synthesis b. Double dis

Bond [772]

Answer: a. Synthesis

Explanation:

a. Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

Example: $Mg+O_2\rightarrow 2MgO$

Thus magnesium in its elemental form is combining with oxygen to form magnesium oxide.

b. Double displacement reaction is one in which exchange of ions take place.

Example: $2NaOH(aq)+(NH_4)_2SO_4(aq)\rightarrow 2NH_4OH(aq)+Na_2SO_4(aq)$

c. Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Example: $Li_2CO_3\rightarrow Li_2O+CO_2$

5 0

1 year ago

Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli

Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>

3 0

2 years ago

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

1 year ago

A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

ladessa [460]

Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Let's call chloroform C and acetone A.

Molar concentration of C = Moles of C/Litres of solution

(a) Moles of C

Assume 0.187 mol of C.

That takes care of that.

(b) Litres of solution

Then we have 0.813 mol of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

If there is no change of volume on mixing.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration = moles of solute/kilograms of solvent

Moles of C = 0.187 mol

Mass of A = 47.22 g = 0.047 22 kg

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

4 0

2 years ago

For the following dehydrohalogenation (E2) reaction, draw the Zaitsev product(s) resulting from elimination involving C3–C4 (i.e

Bond [772]

Answer:

2-methyl-butene

Explanation:

For the E2 mechanism, we have an <u>anti-elimination</u>. The Br leaves the molecule and the base removes the hydrogen in the anti position to form the double that's why only one structure is produced. (See figure 1)

Since we have 2 hydrogens on the right carbon, we cannot indicate a <u>specific stereoisomer</u>, in other words, it is not possible to assign a <u>Z or E</u> configuration for this alkene.

8 0

2 years ago