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adelina 88 [10]

3 years ago

11

What is the mass of insoluble lead(II) iodide (461.0 g/mol) produced from 0.830 g of potassium iodide (166.00 g/mol) and aqueous

lead(II) nitrate?A) 4.61 g. B) 0.149 g.C) 2.31 g.D) 1.15 g.E) 0.598 g.

1 answer:

VikaD [51]3 years ago

7 0

Answer:

C

Explanation:

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Which of the following models a single replacement reaction?

koban [17]

The correct answer would be option 3. I order for a single displacement to occur there must be a single element and one compound. From then, the single element will then swap places with one of the elements in the compound (or in this case single circle and combined circles). Hope this helped!

5 0

3 years ago

The distance between the nuclei of two iron atoms is about 4 angstroms.a. Trueb. False

vladimir2022 [97]

Answer:

The correct option is b. false

Explanation:

The distance between the nucleus of an atom and it's outermost shell is called is atomic radius. The atomic radius of an Iron atom (Fe) is 0.126 nm or 1.26 angstrom. The distance between the nuclei of two Iron atoms will be 1.26 × 2 = 2.52 angstroms.

Since 2.52 angstroms is lower than 4 angstroms, the correct option is false

8 0

3 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

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5 0

2 years ago

if 0.0203 g of a gas dissolves in 1.39 l of water at 1.02 atmospheres at what pressure (in atm) would you be able to dissolve 0.

Sergeu [11.5K]

First, we need to get n1 (no.of moles of water ): when

mass of water = 0.0203 g and the volume = 1.39 L

∴ n1 = mass / molar mass of water

= 0.0203g / 18 g/mol
= 0.00113 moles

then we need to get n2 (no of moles of water) after the mass has changed:

when the mass of water = 0.146 g

n2 = mass / molar mass

= 0.146g / 18 g/ mol
= 0.008 moles

so by using the ideal gas formula and when the volume is not changed:

So, P1/n1 = P2/n2

when we have P1 = 1.02 atm

and n1= 0.00113 moles

and n2 = 0.008 moles

so we solve for P2 and get the pressure

∴P2 = P1*n2 / n1

=1.02 atm *0.008 moles / 0.00113 moles

= 7.22 atm

∴the new pressure will be 7.22 atm

5 0

3 years ago

I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

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:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

3 years ago