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1 year ago

I need help with my homework.Calculate the number of formula units in: 41.5 grams CaBr2

Chemistry

1 answer:

lana66690 [7]1 year ago

5 0

Answer:

$\text{ }1.25\times10^{23}\text{ formula units}$

Explanation:

Here, we want to calculate the number of formula units in the given molecule

We start by getting the number of moles

To get the number of moles, we have to divide the mass given by the molar mass

The molar mass is the mass per mole

The molar mass of calcium bromide is 200 g/mol

Thus, we have the number of moles as follows:

$\frac{41.5}{200}\text{ = 0.2075 mol}$

The number of formula units in a mole is:

$1\text{ mole = 6.02 }\times10^{22}\text{ formula units}$

The number of formula units in 0.2075 mole will be:

$0.2075\text{ }\times6.02\times10^{23}\text{ = }1.25\times10^{23}\text{ formula units}$

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How many electrons would be transferred in either a voltaic or electrolytic cell that uses the following half reactions

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Answer:

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

Explanation:

$Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V$

$Mg^{2+}(aq) + 2 e^- \rightarrow Mg (s),E^o = -2.37 V$

The substance having highest positive reduction $E^o$ potential will always get reduced and will undergo reduction reaction.

Reduction : cathode

$Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V$ ..[1]

Oxidation: anode

$Mg(s)\rightarrow Mg^{2+}(aq) + 2 e^-,E^o = 2.37 V$ ..[2]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

$E^o_{cell}=-0.036V-(-2.37 V)=2.334 V$

The overall reaction will be:

2 × [1] + 3 × [2] :

$2Fe^{3+} (aq) + 3Mg(s)+6e^-\rightarrow 2Fe (s)+3Mg^{2+}(aq)+6e^-$

Electrons on both sides will get cancelled :

$2Fe^{3+} (aq) + 3Mg(s)\rightarrow 2Fe (s)+3Mg^{2+}(aq)$

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

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3 years ago

The use of crucible

Norma-Jean [14]

a ceramic or metal container in which metals or other substances may be melted or subjected to very high temperatures.

"the crucible tipped and the mold filled with liquid metal"

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2 years ago

What does pressure measure

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Answer:

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Explanation:

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3 years ago

Consider the following Reaction.

jeyben [28]

Hey there!:

Molar mass of Mg(OH)2 = 58.33 g/mol

number of moles Mg(OH)2 :

moles of Mg(OH)2 = 30.6 / 58.33 => 0.5246 moles

Molar mass of H3PO4 = 97.99 g/mol

number of moles H3PO4:

moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles

Balanced chemical equation is:

3 Mg(OH)2 + 2 H3PO4 ---> Mg3(PO4)2 + 6 H2O

3 mol of Mg(OH)2 reacts with 2 mol of H3PO4 ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !

Now , we will use Mg(OH)2 in further calculation .

Molar mass of Mg3(PO4)2 = 262.87 g/mol

According to balanced equation :

mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2

= (1/3)*0.5246

= 0.1749 moles of Mg3(PO4)2

use :

mass of Mg3(PO4)2 = number of mol * molar mass

= 0.1749 * 262.87

= 46 g of Mg3(PO4)2

Therefore:

% yield = actual mass * 100 / theoretical mass

% = 34.7 * 100 / 46

% = 3470 / 46

= 75.5%

Hope that helps!

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2 years ago

If a pharmacist adds 10 ml of purified water to 30 ml of a solution having a specific gravity of 1.30, calculate the specific gr

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The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.

Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be $\frac{49}{40}=1.225$ g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.

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