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2 years ago

"35.0 g of silicon dioxide contains how many grams of oxygen? a. 32.0 b. 18.6 c. 1.88 d. 0.530"

Chemistry

1 answer:

Nat2105 [25]2 years ago

7 0

Answer:

b. 18.6

Explanation:

Given that:-

Mass of silicon dioxide = 35.0 g

Molar mass of silicon dioxide = 60.08 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{35.0\ g}{60.08\ g/mol}$

$Moles= 0.58255\ mol$

From the formula, $SO_2$ 1 mole of sulfur dioxide contains 2 moles of oxygen

So,

0.58255 mole of sulfur dioxide contains 2*0.58255 moles of oxygen

moles of oxygen = 1.1651 moles

Molar mass of oxygen = 15.999 g/mol

Mass= Moles*Molar mass = $1.1651\times 15.999\ g$ = 18.6 g

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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

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Answer:

Explanation:

Given that:

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

Pressure = 1800 kPa

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

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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

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6.07 grams is the theoretical yield of calcium phosphate (Ca₃(PO₄)₂).

<h3>How we calculate mass from moles?</h3>

Mass of any substance can be calculated by using moles as:

n = W/M, where

W = required mass

M= molar mass

Given chemical reaction is:

3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO₄)₂

From the stoichiometry it is clear that:

3 moles of Ca(NO₃)₂ = produce 1 mole of Ca₃(PO₄)₂

Given mass of Ca(NO₃)₂ = 96.1g

Mole of Ca(NO₃)₂ = 96.1g/164g/mol = 0.5859moles

So, 0.5859 moles of Ca(NO₃)₂ = produce 0.5859×1/3 = 0.0196 moles of Ca₃(PO₄)₂

Required mass of Ca₃(PO₄)₂ will be calculated by using moles as:

W = 0.0196mole × 310g/mole = 6.07 grams

Hence, 6.07 grams is the theoretical yield of calcium phosphate.

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3 years ago

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