From the middle of the periodic table, in which direction would you look to find the most reactive substances?

A salt forms in the reaction of barium with chlorine. What is the most likely formula unit of this salt?

Scrat [10]

The answer to this question would be: BaCl 2
Barium is an alkali metal with 56 atomic number. Barium located in the group 2 of the periodic table because it has 2 valence electrons. Chlorine is a nonmetal that has 1 valence electron. When react, it would need 2 chlorine for each barium as the valence electron of barium is twice the chlorine.

6 0

3 years ago

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Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing

Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of $AgNO_3$ = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of $AgNO_3$ = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and $AgNO_3$ .

$\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles$

$\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

From the balanced reaction we conclude that

As, 2 mole of $AgNO_3$ react with 1 mole of $Zn$

So, 0.1479 moles of $AgNO_3$ react with $\frac{0.1479}{2}=0.07395$ moles of $Zn$

From this we conclude that, $Zn$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag$

From the reaction, we conclude that

As, 2 mole of $AgNO_3$ react to give 2 mole of $Ag$

So, 0.1479 moles of $AgNO_3$ react to give 0.1479 moles of $Ag$

Now we have to calculate the mass of $Ag$

$\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag$

$\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g$

Theoretical yield of $Ag$ = 15.95 g

Experimental yield of $Ag$ = 13.32 g

Now we have to calculate the percent yield.

$\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100$

$\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%$

Therefore, the percent yield is, 83.51 %

7 0

3 years ago

When light shines on a sample, each element emits specific wavelengths producing a unique fingerprint called its ______ spectra.

DanielleElmas [232]

Answer:

B.

Explanation:

3 0

3 years ago

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Gas has a volume of 247.3 ML and is at 100 Celsius and 745 Hg. If the mass of the gas is 0.347 g what is the molar mass of the v

Ahat [919]

Answer:

The molar mass of the vapor is 43.83 g/mol

Explanation:

Given volume of gas = V = 247.3 mL = 0.2473 L

Temperature = T = 100 $^{\circ}C$ = 373 K

Pressure of the gas = P = 745 mmHg (1 atm = 760 mmHg)

$P = \displaystyle \frac{745}{760} \textrm{ atm} = 0.9802 \textrm{ atm}$

Mass of vapor = 0.347 g

Assuming molar mass of gas to be M g/mol

The ideal gas equation is shown below

$\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\0.98026 \textrm{ atm}\times 0.2473 \textrm{ L} = \displaystyle \frac{3.47 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 373\textrm{K} \\M = 43.834 \textrm{ g/mol}$

The molar mass of the vapor comes out to be 43.834 g/mol

4 0

2 years ago

What are two versions of genes called and how are they diffrent?

Stolb23 [73]

Answer with Explanation:

Different versions of a gene are called alleles. Alleles are described as either dominant or recessive depending on their associated traits. Since human cells carry two copies of each chromosome they have two versions of each gene.

8 0

2 years ago