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tamaranim1 [39]

2 years ago

8

iron combines with 4.00 g of copper (11) nitrate to form 6.01 g of Iron (l) nitrate and 0.400 g copper metal. how much iron did

it take to convert to Cu(NO3)2?

1 answer:

kvasek [131]2 years ago

7 0

Answer:

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Explanation:

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Which of the following species is not formed through a termination reaction in the chlorination of methane? Which of the followi

krek1111 [17]

Explanation:

When we add chlorine to a substance or compound then this process is known as chlorination.

For example, a process of chlorination is as follows.

Initiation : $Cl_{2} \overset{light}{\rightarrow} 2Cl*$

where, Cl* is a free radical.

Propagation:

$Cl* + CH_{4} \rightarrow HCl + *CH_{3}$

$CH_{3} + Cl_{2} \rightarrow CH_{3}Cl + Cl*$

Termination:

$2Cl* \rightarrow Cl_{2}$

$Cl* + *CH_{3} \rightarrow CH_{3}Cl$

$2*CH_{3} \rightarrow CH_{3}-CH_{3}$

Thus, we can conclude that out of the given options $H_{2}$ is not formed through a termination reaction in the chlorination of methane.

5 0

2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

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

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

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

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

2 years ago

A student titrated 25.0 cm3 portions of dilute sulfuric acid with a 0.105 mol/dm3 sodium hydroxide solution.The equation for the

Rzqust [24]

Answer:

This question is incomplete

Explanation:

This question is incomplete as the volume of the base that was used during the titration was not provided. However, the completed question is in the attachment below.

The formula to be used here is CₐVₐ/CbVb = nₐ/nb

where Cₐ is the concentration of the acid = unknown

Vₐ is the volume of the acid used = 25 cm³ (as seen in the question)

Cb is the concentration of the base = 0.105 mol/dm³ (as seen in the question)

Vb is the volume of the base = 22.13 cm³ (22.1 + 22.15 + 22.15/3)

nₐ is the number of moles of acid = 1 (from the chemical equation)

nb is the number of moles of base = 2 (from the chemical equation)

Note that the Vb was based on the concordant results (values within the range of 0.1 cm³ of each other on the table) of the student

Cₐ x 25/0.105 x 22.13 = 1/2

Cₐ x 25 x 2 = 0.105 x 22.13 x 1

Cₐ x 50 = 0.105 x 22.13

Cₐ = 0.105 x 22.13/50

Cₐ = 0.047 mol/dm³

The concentration of the sulfuric acid is 0.047 mol/dm³

7 0

1 year ago

List the following compounds in decreasing electronegativity difference. cl2 hcl nacl

Mkey [24]

Based on Pauling Scale, electro negativity of Cl = 3.2, Na = 0.9 and H = 2.1

Thus, Electronegativity difference  in $Cl_{2}$ = 3.2 -3.2 = 0
Electronegativity difference  in NaCl = 3.2-0.9 = 2.3
Similarly, Electronegativity difference  in HCl = 3.2 - 2.1 = 1.1

Thus, among the listed molecules following is the decreasing order of electronegativity difference: NaCl> HCl > $Cl_{2}$

8 0

2 years ago

1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theore

Alina [70]

Answer :

(1) The theoretical yield of product, $MgO$ is, 1.257 grams.

(2) The percent yield of $MgO$ is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

$\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles$

Now we have to calculate the moles of $MgO$

The balanced chemical reaction is,

$2Mg(s)+O_2(g)\rightarrow 2MgO(s)$

From the reaction, we conclude that

As, 2 mole of $Mg$ react to give 2 mole of $MgO$

So, 0.03142 moles of $Mg$ react to give 0.03142 moles of $MgO$

Now we have to calculate the mass of $MgO$

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

$\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g$

Theoretical yield of $MgO$ = 1.257 g

Experimental yield of $MgO$ = 0.8922 g

Now we have to calculate the percent yield of $MgO$

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

$\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%$

Therefore, the percent yield of $MgO$ is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

4 0

2 years ago