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2 years ago

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The force of attraction between a divalent cation and a divalent anion is 1.91 x 10-8 n. If the ionic radius of the cation is 0.

074 nm, what is the anion radius?

1 answer:

Masteriza [31]2 years ago

7 0

The force of attraction (F) is given by the formula:

F = (1/4π∈r²)(Zc*e)(Za*e)---------(1)

where:

∈ = permittivity of free space = 8.85*10^-15 F/m

Zc = charge on the cation = +2

Zc = charge on the anion = -2

e = charge on an electron = 1.602*10^-19 C

r = interionic distance = rc + ra

where rc and ra are the radius of cation and anion respectively

F = 1.91 * 10^-8 N

therefore based on equation (1) we have:

1.91 *10^-8 = [1/4π(8.85*10^-15)r²](2*1.602*10^-19)²

r² = 0.04832 * 10^-15

r = 6.951 nm

Now: r = rc + ra

where rc = 0.074 nm

thus, ra = r - rc = 6.951 - 0.074 = 6.877 nm

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Just Lemons Lemonade Recipe Equation:

zalisa [80]

Answer:

Explanation:

Hello!

<em>Complete text:</em>

<em>Honors Stoichiometry Activity WorksheetInstructions: </em>

<em>Activity Two: Just Lemons, Inc. Production</em>

<em>Here's a one-batch sample of Just Lemons lemonade production. Determine the percent yield and amount of leftover ingredients for lemonade production and place your answers in the data chart.</em>

<em>Hint: Complete stoichiometry calculations for each ingredient to determine the theoretical yield. Complete a limiting reactant-to-excess reactant calculation for both excess ingredients. </em>

<em>Water 946.36 g </em>

<em>Sugar 196.86 g </em>

<em>Lemon Juice 193.37 g </em>

<em>Lemonade 2050.25g</em>

<em>Leftover Ingredients?</em>

<em>Just Lemons Lemonade Recipe Equation:</em>

<em>2 water + sugar + lemon juice = 4 lemonade</em>

<em>Mole conversion factors:</em>

<em>1 mole of water = 1 cup = 236.59 g</em>

<em>1 mole of sugar = 1 cup = 225 g</em>

<em>1 mole of lemon juice = 1 cup = 257.83 g</em>

<em>1 mole of lemonade = 1 cup = 719.42 g</em>

You have the information on the ingredients used to produce one batch of lemonade and the amount of lemonade produced. To determine which ingredients be leftovers, you have to determine first, which one is the limiting reactant, i.e. the ingredient that will be used up first.

According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g = 4*(719.42)g

473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

<u>Water:</u>

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

<u>Sugar:</u>

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

<u>Lemon juice:</u>

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

I hope this helps!

7 0

2 years ago

How many moles of O2 should be supplied to burn 1 mol of C3H8 (propane) molecules in a camping stove

Shkiper50 [21]

The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Each mole of propane requires 5 moles of oxygen.

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2 years ago

Based on the crystal-field strengths cl- < f- < h2o < nh3 < h2nc2h4nh2, which octahedral ti (iii) complex below has

kati45 [8]

<span>Based on the crystal field strength, Cl ligand would give the longest d-d transition when complexed with Ti(III). as this is the weak field ligand and would cause minimum splitting of d orbitals.</span>

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2 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago

A gas occupies 72.1 at stp. At what temperature would the gas occupy 85.9 L at a pressure of 93.6 kPa?

Alika [10]

Answer:

328.1 K.

Explanation:

To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT</em>.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in.

If n is constant, and have two different values of (P, V and T):

<em>P₁V₁T₂ = P₂V₂T₁</em>

<em></em>

P₁ = 1.0 atm (standard P), V₁ = 72.1 L, T₁ = 25°C + 273 = 298 K (standard T).

P₂ = 93.6 kPa = 0.924 atm, V₂ = 85.9 L, T₂ = ??? K.

<em>T₂ = P₂V₂T₁/P₁V₁ = </em>(0.924 atm)(85.9 L)(298 K)/(1.0 atm)(72.1 L) <em>= 328.1 K.</em>

<em></em>

4 0

2 years ago