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2 years ago

Write the common names of trichloromethane and one of their uses?

Chemistry

1 answer:

ankoles [38]2 years ago

8 0

Answer:

One is chloroform, formerly used as an inhaled anesthetic during surgery, the primary use of chloroform today is in industry, where it is used as a solvent and in the production of the refrigerant freon.

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Can you help me balance the equations from 1-10 please and thank you.

Korolek [52]

Answer:

2 NaBr + Ca(OH)2 → CaBr2 + 2NaOH
2 NH3 +H2SO4 → (NH4)2(SO4)

8 0

2 years ago

If i have 8 L of 0.25 M solution, how many miles of MgCl2 are present

qaws [65]

Answer:

2 moles

Explanation:

The following were obtained from the question:

Molarity = 0.25 M

Volume = 8L

Mole =?

Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:

Molarity = mole of solute/Volume of solution.

With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:

Molarity = mole of solute/Volume of solution.

0.25 = mole of MgCl2 /8

Cross multiply to express in linear form

Mole of MgCl2 = 0.25 x 8

Mole of MgCl2 = 2 moles

Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution

4 0

2 years ago

How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with

konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = $a=31 cm^2$

Thickness of the gold plating = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

$V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3$

Density of the gold = $d=19.3 g/cm^3$

$m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g$

Moles of gold = $\frac{29.915 g}{197 g/mol}=0.152 mol$

$Au^{3+}+3e^-\rightarrow Au$

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

$\frac{3}{1}\times 0.152 mol=0.456 mol$ of electrons

Number of electrons = N = $0.456\times \times 6.022\times 10^{23}$

Charge on single electron = $q=1.6\times 10^{-19} C$

Total charge required = Q

$Q=N\times q$

Amount of current passes = I = 8 Ampere

Duration of time = T

$I=\frac{Q}{T}$

$T=\frac{N\times q}{I}$

$=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s$

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

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2 years ago

Am i the only one getting a red text saying my question is mean?

Triss [41]

Answer:

I dont know

Explanation:

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2 years ago

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Sladkaya [172]

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2 years ago

Write the common names of trichloromethane and one of their uses?​

Write the common names of trichloromethane and one of their uses?