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Mazyrski [523]
4 years ago
15

What is the mass of 1.23 mol NaCl? Please explain

Chemistry
1 answer:
storchak [24]4 years ago
6 0
Look up a periodic table and add up the molar mass of NaCl.
Na is 11g/mol
Cl is 17 g/mol
NaCl is 11+17=28 g/mol

1.23 mol NaCl * 28 g/mol NaCl = 34.44 g NaCl
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How do you do empirical formula
mihalych1998 [28]

Answer:

ez

Explanation:

Step 1: Obtain the mass of each element present in grams. Element % = mass in g = m.

Step 2: Determine the number of moles of each type of atom present. ...

Step 3: Divide the number of moles of each element by the smallest number of moles. ...

Step 4: Convert numbers to whole numbers.

7 0
3 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
3 years ago
Someone please help i don’t have much time left
REY [17]

Answer: Energy of reactants = 30, Energy of products = 10

Exothermic

Activation energy for forward reaction is 10.

Explanation:

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and \Delta H for the reaction comes out to be negative.

Energy of reactants = 30

Energy of products = 10

Thus as energy of the product < energy of the reactant, the reaction is exothermic.

Activation energy (E_a) is the extra energy that must be supplied to reactants in order to cross the energy barrier and thus convert to products.

E_a for forward reaction is (40-30) = 10.

6 0
3 years ago
Freezing point<br> Where would you place this property in the Venn diagram?
Marat540 [252]
For the Venn diagram, freezing water should be in the physical category (A)
7 0
3 years ago
Read 2 more answers
9) After lab, all of Darrel’s friends looked at his data and laughed and laughed. They told him that he was 30.8% too low in the
zaharov [31]
If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.

In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.

So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC
3 0
4 years ago
Read 2 more answers
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