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3 years ago

Aqueous sodium hydroxide forms a light blue precipitate. What is the formula of the blue precipitate?

Chemistry

1 answer:

lisabon 2012 [21]3 years ago

8 0

Answer:

Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)

Explanation:

Use of aqueous sodium hydroxide is a precipitation reaction to test for anions or cations. Aqueous sodium hydroxide in a precipitate test forms a insoluble precipitates along with some colors characteristics.

Aqueous sodium hydroxide (NaOH) when mixed with copper(II) (Cu2+) forms a blue precipitate. The formula is as follows:

Cu2+(aq) + 2OH-(aq) => Cu(OH)2(s)

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How many moles are there in 270.4 g sample of NaOH

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6.5

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No of moles is the ratio of reacting mass to molar mass

RM - 270.4g

MM - 23+16+1= 40g/mol

270.4/40= 6.76mol

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2 years ago

Nickel carbonyl, Ni(CO)4 is one of the most toxic substances known. The present maximum allowable concentration in laboratory ai

vodomira [7]

<u>Answer:</u> The mass of $Ni(CO)_4$ allowable in the laboratory is 4599.5 grams

<u>Explanation:</u>

To calculate the volume of cuboid, we use the equation:

$V=l\times b\times h$

where,

V = volume of cuboid

l = length of cuboid = 14 ft

b = breadth of cuboid = 22 ft

h = height of cuboid = 9 ft

Putting values in above equation, we get:

$V=14\times 22\times 9=2772ft^3=78503.04L$ (Conversion factor: $1ft^3=28.32L$

To calculate the moles of gas, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 78503.04 L

T = Temperature of the gas = $23^oC=[23+273]K=296K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of hydrogen gas = ?

Putting values in above equation, we get:

$1.00atm\times 78503.04L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 296K\\\\n=\frac{1.00\times 78503.04}{0.0821\times 296}=3230.4mol$

Applying unitary method:

For every 109 moles of gas, the moles of $Ni(CO)_4$ present are 1 moles

So, for 3230.4 moles of gas, the moles of $Ni(CO)_4$ present will be = $\frac{1}{109}\times 3230.4=26.94mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of $Ni(CO)_4$ = 26.94 moles

Molar mass of $Ni(CO)_4$ = 170.73 g/mol

Putting values in above equation, we get:

$26.94mol=\frac{\text{Mass of }Ni(CO)_4}{170.73g/mol}\\\\\text{Mass of }Ni(CO)_4=(26.94mol\times 170.73g/mol)=4599.5g$

Hence, the mass of $Ni(CO)_4$ allowable in the laboratory is 4599.5 grams

3 0

3 years ago