Is there a relationship between the number of black boxes and any of the water pumps

5. calculate the percentage of water of crystallization in blue vitriol (copper sulphate pentahydrate), cuso4.5h2o

ki77a [65]

The percentage of water of crystallization in blue vitriol is 36.07%.

M(H₂O) = 2Ar(H) + Ar(O) x g/mol

M(H₂O) = 2 + 16 x g/mol

M(H₂O) = 18 g/mol; molar mass of water

M(CuSO₄·5H₂O) = Ar(Cu) + Ar(S) + 4Ar(O) + 5Mr(H₂O) x g/mol

M(CuSO₄·5H₂O) = 63.5 + 32 + 64 + 90 x g/mol

M(CuSO₄·5H₂O) = 249.5 g/mol; molar mass of copper sulphate pentahydrate

The percentage of water: 5M(H₂O) / M(CuSO₄·5H₂O) x 100%

The percentage of water: 90 g/mol / 249.5 g/mol x 100%

The percentage of water = 36.07%

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5 0

1 year ago

Read 2 more answers

A 40.2 g sample of a metal heated to 99.3Â°C is placed into a calorimeter containing 120 g of water at 21.8Â°C. The final temper

aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

$m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)$

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,

$120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)$

$40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184$

$40.2C_{metal}\left(99.3-24.5\right)=1355.616$

$C_{metal}=0.45\ J/g^0C$

<u>This value corresponds to iron. Thus answer is B.</u>

3 0

3 years ago

onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀

Sveta_85 [38]

Answer:

9.34x10^-4

Explanation:

Step 1:

The balanced equation for the reaction.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

Step 2:

Data obtained from the question:

Mass of PbCl2 = 0.2393 g

Volume = 50mL

concentration of Pb^2+, [Pb^2+] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant, Kc =?

Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

Molar Mass of PbCl2 = 207 + (35.5x2) = 278g/mol

Mass of PbCl2 = 0.2393 g

Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.61x10^-4/0.05

Molarity of PbCl2 = 0.01722 M

Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

[Pb^2+] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

5 0

2 years ago

Calculate the molarity of a solution prepared by dissolving 8.054 g of k2cro7 in enough water to give 250. ml of solution.

Thepotemich [5.8K]

The answer should be 0.077M

7 0

2 years ago

A solution of K2SO4 and KCl is added to a solution of Ba(NO3)2. Which of these compounds will precipitate out of this combined s

Archy [21]

A solution of K2SO4 and KCl is added to a solution of Ba(NO3)2. BaS0₄ (s) will precipitate out of this combined solution.

Molecular equation

K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2 KNO3(aq)?

This equation represents a double displacement (replacement) reaction, also called a metathesis reaction, in which the reactant ions exchange places to form new products. The general equation is:

A-B + C-D → A-D + C-B;

where A and C are cations, and B and D are anions.

Complete ionic equation: Includes all ions and the precipitate.

2K^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2[NO3]^-(aq) → 2K^+(aq) + 2[NO3]^- + BaSO4(s)

In an aqueous solution, precipitation is the process of transforming a dissolved substance into an insoluble solid from a super-saturated solution.

The solid formed is called the precipitate. In case of an inorganic chemical reaction leading to precipitation, the chemical reagent causing the solid to form is called the precipitant.

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6 0

10 months ago