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mixas84 [53]
3 years ago
6

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

Explanation:

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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Tres diferencias entre los líquidos y los sólidos
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When aqueous solutions of __________ are mixed, a precipitate forms.
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nibr2 and agno3

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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
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