Ask question

Ask question

All categories

3 years ago

11

A white salt containing an unknown metal has the formula MCl and gives a lilac flame during a flame test. the salt could be

1 answer:

Tatiana [17]3 years ago

7 0

Answer:

C

Explanation:

KCl - the flame test for Potassium produces a lilac flame

You might be interested in

An egg sinks in fresh water but it floats on salty water.why?

Sonbull [250]

The salt makes it rise and

8 0

3 years ago

Compound A reacts with Compound B to form only one product, Compound C, and it's known the usual percent yield of C in this reac

frez [133]

Given :

Compound A reacts with Compound B to form only one product, Compound C.

The usual percent yield of C in this reaction is 40%.

10.0 g of A are reacted with excess Compound B, and 6.4 g of Compound C

To Find :

The theoretical yield of C.

Solution :

We know, % yield is given by :

$\%\ yield = \dfrac{actual\ yield}{theoretical\ yield }\times 100$

Putting given values , we get :

$40 = \dfrac{6.4}{theoretical\ yield }\times 100\\\\theoretical\ yield=\dfrac{6.4\times 100}{40}\\\\theoretical\ yield=16\ g$

Therefore, theoretical yield of C is 16 g.

Hence, this is the required solution.

7 0

3 years ago

The balanced equation below shows the products that are formed when butane (C4H10) is combusted.

Nataly [62]

Answer:

2:8

Explanation:

The reaction equation is a given as:

2C₄H₁₀ + 130₂ → 8CO₂ + 10H₂O

From the reaction equation, the mole ratio is 2:8

Butane is C₄H₁₀

Carbon dioxide CO₂

From the reaction;

2 moles of butane will produce 8 moles of carbon dioxide

3 0

3 years ago

Read 2 more answers

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

4. Which of the following NOT TRUE about real gas

Semenov [28]

Answer:

The answer is D.

Explanation:

Intermolecular force are negligible

When the distance between molecules decrease,

the attraction or repulsion become greater

7 0

2 years ago