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2 years ago

If the sample has a total mass of 5.76 g and contains 1.79 g k, what are the percentages of kbr and ki in the sample by mass

Chemistry

1 answer:

Luden [163]2 years ago

8 0

The percentages of KBr and KI in the sample by mass is 80.68 and 19.32 % respectively.

<h3>What is Molar Mass ?</h3>

Molar mass is defined as the mass contained in 1 mole of sample.

It is given that

the sample has a total mass of 5.76 g contains KBr and KI

and contains 1.79 gm of K is present

what is the percentage of KBr and KI in the sample

Molecular weight of K = 39

Molecular weight of Br = 79.9

Molecular weight of I = 126.9

In KBr the mass percentage of K is 39/(39+79.9) = 32.89%

In KI the mass percentage of K is 39/(39+126.9) = 23.5%

Let the mass of KBr present in the sample is x

K will be 0.3289 x

and let the mass of KI present be y

K will be 0.235y

x +y =5.76

0.3289x+0.235y = 1.79

0.0939y = 0.1045

y = 1.1125 gm

x = 5.76-1.1125

x = 4.6475 gm

% of KBr = (4.6475/5.76 )*100 = 80.68 %

% of KI = (1.1125/5.76) *100 = 19.32%

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Nezavi [6.7K]

Explanation:

For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count.

a. 20.0 meters x 0012.65 meters

The numbers are;

20.0 (3 s.f) and 12.65 (4 s.f)

The multiplication gives; 253

Since 253 is already in 3 s.f, that's the answer.

b. 002.5 × 103 meters + 3.50 × 102 meters

The numbers are;

002.5 (2 s.f), 103 (3 s.f), 3.50 (3 s.f) and 102 (3 s.f)

002.5 × 103 = 257.5 = 260 (2 s.f)

3.50 × 102 = 357 = 360 (2 s.f)

260 + 360 = 620

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3 years ago

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

<u>Answer:</u> The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

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3 years ago

2.... What must all organic compounds contain? H-O-P <br> 0-S-N <br> C-H-O <br>K-O-H

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Answer:

C-H-O is the answer for this one

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4 years ago

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A reaction proceeds with 2.72 moles of magnesium chlorate and 3.14 moles of sodium hydroxide. this is the equation of the reacti

den301095 [7]

The theoretical amount of each product obtained are:

1.57 moles of Mg(OH)₂
3.14 moles of NaClO₃

<h3>Balanced equation</h3>

Mg(ClO₃)₂ + 2NaOH —> Mg(OH)₂ + 2NaClO₃

<h3>How to determine the limiting reactant</h3>

From the balanced equation above,

1 mole of Mg(ClO₃)₂ reacted with 2 moles of NaOH

Therefore,

2.72 moles of Mg(ClO₃)₂ will react with = 2.72 × 2 = 5.44 moles of NaOH

From the calculation above, we can see that a higher amount (5.44 moles) of NaOH than what was given (3.14 moles) is needed to react completely with 2.72 moles of Mg(ClO₃)₂

Therefore, NaOH is the limiting reactant

<h3>How to determine the theoretical yield of Mg(OH)₂</h3>

From the balanced equation above,

2 moles of NaOH reacted to produce 1 mole of Mg(OH)₂

Therefore,

3.14 moles of NaOH will react to produce = 3.14 / 2 = 1.57 moles of Mg(OH)₂

Thus, the theoretical yield of Mg(OH)₂ is 1.57 moles

<h3>How to determine the theoretical yield of NaClO₃</h3>

From the balanced equation above,

2 moles of NaOH reacted to produce 2 mole of NaClO₃

Therefore,

3.14 moles of NaOH will also react to produce 3.14 moles of NaClO₃

Thus, the theoretical yield of NaClO₃ is 3.14 moles

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2 years ago

This man is using a block and tackle to lift a heavy bale of cotton. the block and tackle contains which simple machine?

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The simple machine used is called atwood machine.

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