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mihalych1998 [28]
3 years ago
15

Calculate the energy required to produce 7.00 mol Cl2O7 on the basis of the following balanced equation. 2Cl2(g) + 7O2(g) + 130

kcal --> 2Cl2O7(g) Select one: a. 7.00 kcal b. 65 kcal c. 130 kcal d. 455 kcal
Chemistry
1 answer:
KatRina [158]3 years ago
3 0

Explanation:

As the given chemical reaction equation is as follows.

      2Cl_{2}(g) + 7O_{2}(g) + 130 kcal \rightarrow 2Cl_{2}O_{7}(g)

Also, it is given that for 2 moles the energy required is 130 kcal. This means that energy required for 1 mole is calculated as follows.

                   1 mole = \frac{130 kcal}{2}

                               = 65 kcal

Hence, energy required for 7 moles will be calculated as follows.

              Energy required = 7 \times 65 kcal

                                           = 455 kcal

Thus, we can conclude that energy required to produce 7.00 mol Cl_{2}O_{7} on the basis of given reaction is 455 kcal.

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Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a dist
strojnjashka [21]

Answer:

Force of attraction = 35.96 \times 10^{27}N

Explanation:

Given: charge on anion = -2

Charge on cation = +2

Distance = 1 nm = 10^{-9} m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

F = \dfrac{k \times q_1 q_2}{ \r^2} ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = 8.99 \times 10^9 \ Nm^{2}C^{-2}

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }

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6 0
2 years ago
According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 gra
Ugo [173]

Answer:

There is 37.36 grams of K3PO4 produced

Explanation:

Step 1: Data given

Mass of H3PO4 = 29.6 grams

KOH is in excess

Molar mass of KOH = 56.11 g/mol

Molar mass of H3PO4 = 97.99 g/mol

Step 2: The balanced equation

3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)

Step 3: Calculate mass of KOH

Mass KOH = mass KOH / molar mass KOH

Mass KOH = 29.6 grams / 56.11 g/mol

Mass KOH = 0.528 moles

Step 4: Calculate moles of K3PO4

Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O

For 0.528 moles of KOH we'll have 0.528/3 =  0.176 moles of K3PO4

Step 5: Calculate mass of K3PO4

Mass K3PO4 = moles K3PO4 * molar mass K3PO4

Mass K3PO4 = 0.176 moles * 212.27 g/mol

Mass K3PO4 = 37.36 grams

There is 37.36 grams of K3PO4 produced

8 0
3 years ago
The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The
dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

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Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

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Answer:

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