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3 years ago

Which element would mostly have chemical properties similar to that of magnesium (Mg)?

1 answer:

7 0

The elements of same group are placed in same group as they have similar properties. So Calcium is the element which resembles Mangesium.

Calcium and Magnesium both are alkaline earth metals.

They have similar properties as they have similar outer or valence shell electronic configuration

The general electronic configuration of alkaline earht metal is ns2

The electronic configuration of Mg [atomic number 12] and Ca [atomic number 20] are

$[Ne]3s^{2}$ and $[Ar]4s^{2}$ respectively.

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PLS PLS PLS PLS HELP IM STRUGGLING

andriy [413]

Answer:

A. 4 only

Explanation:

One of the tests for purity is by determining the melting point of a substance. A pure substance has a sharp meting point.

If the solid in question is a pure solid, when it is heated, it will exhibit a sharp melting point. Mixtures however melt over a range of temperatures due to the presence of impurities in the solid.

5 0

3 years ago

Calculate the percent yield of benzyl alcohol Sue obtained when doing the following reaction. Sue started with 2.1 g of benzoic

Pachacha [2.7K]

Answer:

The percent yield is 116.6 %

Explanation:

Step 1: Data given

Mass of benzoic anhydride = 2.1 grams

Mass of sodium borohydride = 0.15 grams

Molar mass of benzoic anhydride = 226.23 g/mol

Molar mass of NaBH4 = 37.83 g/mol

Mass of benzyl alcohol produced = 0.5 grams

Step 2: The balanced equation

C14H10O3 + NaBH4 → C7H6O2 + C7H8O + NaB

Step 3: Calculate moles of C14H10O3

Moles C14H10O3 = Mass / molar mass

Moles C14H10O3 = 2.10 grams / 226.23 g/mol

Moles C14H10O3 = 0.00928 moles

Step 4: Calculate moles NaBH4

Moles NaBH4 = 0.150 grams / 37.83 g/mol

Moles NaBH4 = 0.00397 moles

Step 5: Calculate limiting reactant

For 1 mol of C14H10O3 we need 1 mol of NaBH4 to produce 1 mol of C7H8O

NaBH4 is the limiting reactant. It will completely be consumed ( 0.00397 moles).

C14H10O3 is in excess. There will react 0.00397 moles.

There will remain 0.00928 - 0.00397 = 0.00531 moles

Step 6: Calculate moles of C7H8O

For 1 mol of C14H10O3 we need 1 mol of NaBH4 to produce 1 mol of C7H8O

For 0.00397 of C14H10O3 we need 0.00397 mol of NaBH4 to produce 0.00397 mol of C7H8O

Step 7: Calculate mass of C7H8O

Mass of C7H8O = Moles * molar mass

Mass of C7H8O = 0.00397 moles * 108.14 g/mol

Mass of C7H8O = 0.429 grams = theoretical yield

Step 8: Calculate % yield

% yield = (actual yield/ theoretical yield) * 100%

% yield = 0.5/0.429) *100 %

% yield = 116.6 %

The percent yield is 116.6 %

3 0

4 years ago

Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl

Cerrena [4.2K]

Answer:

0.077 M

Explanation:

Molarity is the representation of the solution.

Molarity:

It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

M = moles of solute / Liter of solution . . . . . . . . . . (1)

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in L = 100 mL / 1000

Volume of Solution = 0.1 L

molar mass of NaCl = 58.44 g/mol

Now to find number of moles of Nacl

no. of moles of NaCl = mass of NaCl / molar mass

no. of moles of NaCl = 0.45g / 58.44 g/mol

no. of moles of NaCl = 0.0077 g

Put values in the eq (1)

M = moles of solute / Liter of solution . . . . . . . . . . (1)

M = 0.0077 g / 0.1 L

M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

3 0

4 years ago

Iodine is prepared both in the laboratory and commercially by adding Cl2(g) to an aqueous solution containing sodium iodide. 2Na

otez555 [7]

Answer:

79.0 g

Explanation:

1. Gather the information in one place.

MM: 148.89 253.81

2NaI + Cl2 → I2 + 2NaCl

m/g: 67.3

2. Moles of I2

n = 67.3 g × (1 mol/253.81 g) = 0.2652 mol I2

2. Moles of NaI needed

From the balanced equation, the molar ratio is 2 mol NaI: 1 mol I2

n = 0.028 76 mol I2× (2 mol NaI/1 mol I2) = 0.5303 mol NaI

3. Mass of NaI

m = 0.5303 mol × (148.89 g/1 mol) = 79.0 g NaI

It takes 79.0 g of NaI to produce 67.3 g of I2.

6 0

4 years ago

Identify the term that matches each electrochemistry definition.

anygoal [31]

Answer: a. Cathode

b. Galvanic cell

c. Anode

d. Electrolytic cell

e. half reaction

Explanation:

Galvanic cell or Electrochemical cell is defined as a device which is used for the conversion of the chemical energy produced in a spontaneous redox reaction into the electrical energy.

Electrolytic cell is a device where electrical energy is used to drive a non spontaneous chemical reaction.

In the electrochemical cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode. Thus the electrons are produced at anode and travel towards cathode.

The balanced two-half reactions will be:

Oxidation half reaction : $M\rightarrow M^{n+}+ne^-$

Reduction half reaction : $N^{n+}+ne^-\rightarrow N$

Thus the overall reaction will be: $M+N^{n+}\rghtarrow M^{n+}+N$

4 0

3 years ago