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2 years ago

13

Which method is adopted in the seperation of lead chloride from water

1 answer:

Sergio039 [100]2 years ago

3 0

Explanation:

by dissolving the mixture of lead sulphate and lead chloride in water we can separate the two. after dissolving the mixture on water lead sulphate can be obtained as the solid that 's left behind lead chloride can be recovered by evaporating.

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the size of the negative charge of an electron us exactly the same as the size of the positive charge of a proton. what is the o

Ket [755]

Surrounded by a cloud of negatively charged electrons The atomic nucleus consists of positively charged protons and electrically neutral neutrons. (except in the case of Hydrogen-1, which is
154 KB (12,134 words) - 15:57, April 9, 2021

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2 years ago

A sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C^−1 , is put into a calorimeter (see sketch at right) t

Tcecarenko [31]

Answer: The mass of aluminium sample is 55.4 gram

Explanation:

$Q_{absorbed}=Q_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$

where,

$m_1$ = mass of aluminium = ?

$m_2$ = mass of water = 300.0 g

$T_{final}$ = final temperature = $23.8^0C$

$T_1$ = temperature of aluminium = $94.5^oC$

$T_2$ = temperature of water = $21.0^oC$

$c_1$ = specific heat of aluminium = $0.897J/g^0C$

$c_2$ = specific heat of water = $4.184J/g^0C$

Now put all the given values in equation (1), we get

$-[m_1\times c_1\times (T_{final}-T_1)]=[m_2\times c_2\times (T_{final}-T_2)]$

$-[m_1\times 0.897\times (23.8-94.5)^0C]=[300.0g\times 4.184\times (23.8-21.0)]$

$m_1=55.4g$

Therefore, the mass of aluminium sample is 55.4 gram

8 0

3 years ago

How many significant figures does 6559.060 have

UNO [17]

<span>significant figures: 7, decimals: 3</span>

5 0

3 years ago

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

2 years ago

A city in Laguna generates 0.96 kg per capita per day of Municipal Solid Waste (MSW). Makati City in Metro Manila generates 1.9

mamaluj [8]

Answer:

i) amount of MSW generated:

Laguna: 19200 Kg/day

Makati: 38000 Kg/day

ii) number of trucks to collect twice weekly:

Laguna: 3 trucks

Makati: 5 trucks

iii) volume of MSW in tons that enter landfill/week:

Laguna: 147.84 ton/week

Makati: 292 ton/week

Explanation:

i) Laguna: 0.96 Kg person / day of MSW * 20000 = 19200 Kg MSW / day

⇒ Laguna: 19200 Kg/day * ( 7day/ week ) = 134400 Kg/week

Makati: 1.9 Kg person / day of MSW * 20000 = 38000 Kg MSW / day

⇒ Makati: 38000 Kg/day * ( 7day/week ) = 266000 Kg/week

ii) truck capacity = 4.4 ton * ( Kg / 0.0011 ton ) = 4000 Kg

⇒ quote/day = 4000 Kg * 0.75 = 3000 Kg

⇒ loads/day = 2 * 3000 kg = 6000 Kg

⇒ operate/week = 5 * 6000 Kg = 30000 Kg

∴ Laguna: number of trucks needed/week= 134400 / 30000 = 4.48 ≅ 5 trucks

⇒ number of trucks to collect twice weekly = 5 / 2 = 2.5 ≅ 3 trucks

∴ Makati : number of trucks needed/week = 266000 / 30000 = 8.86 ≅ 9 trucks

⇒ number of trucks to collect twice weekly = 9 / 2 = 4.5 ≅ 5 trucks

iii) enter landfill/week:

Laguna: 134400Kg MSW/week * ( 0.0011 ton/Kg ) = 147.84 ton/week

Makati: 266000Kg MSW/week * ( 0.0011 ton/Kg ) = 292 ton/week

4 0

3 years ago