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horrorfan [7]
3 years ago
5

Did the test solution react with the supernatant from the test tube that had the largest amount of precipitate? what can you inf

er about the molar ratio of this reaction?
Chemistry
1 answer:
nydimaria [60]3 years ago
5 0

Answer: -

The test solution reacted with the supernatant from the test tube. In trial number 3 there was the largest amount of precipitate.

An amount of 0.019 g of precipitate was found. In trial 3, 3 mL CuSO₄ and 3 mL Na₂S was taken. Thus they were at a 1: 1 ratio.

From the fact that the most amount of precipitate was formed when the two reactants were at 1:1 ratio implies that at that ratio, all of the reactants reacted.

Thus the mole ratio of the reaction is 1 : 1.

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For water at 30C and 1 atm: a= 3.04x10^-4 k^-1 , k =4.52x10^-5 atm ^-1 = 4.46 x10^-10 m^2/N, cpm= 75.3j/9molk), Vm =18.1cm^3/mol
Lapatulllka [165]

Answer:

C_{vm} of water at 30C and 1 atm is 256.834 J/mol·K.

Explanation:

To solve the question, we note the Maxwell relation such as

C_{pm}-C_{vm}=\frac{9T\alpha ^2 V }{K}

Where:

C_{pm} = Specific heat of gas at constant pressure = 75.3 J/mol·K

C_{vm} = Specific heat of gas at constant volume = Required

T = Temperature = 30 °C = 303.15 K

α = Linear expansion coefficient = 3.04 × 10⁻⁴ K⁻¹

K = Volume comprehensibility = 4.52 × 10⁻⁵ atm⁻¹

Therefore,

75.3 - C_v = \frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81  \times 10^{-5}  }{4.52 \times 10^{-5} }

C_{vm}  = \frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81  \times 10^{-5}  }{4.52 \times 10^{-5} } - 75.3 = 256.834 J/mol·K.

8 0
3 years ago
In ocean water, salt is a(n) _____.<br><br> a. solution<br> b. alloy<br> c. solvent<br> d. solute
stepan [7]
The answer is a. solution :)
3 0
3 years ago
Read 2 more answers
Identify the type of molecule shown in the picture
BlackZzzverrR [31]
1) acid
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7 0
3 years ago
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
4 years ago
Rusting of iron can be prevented by wrapping iron articles in newspaper true or false​
pantera1 [17]

false, the rusting of iron can be prevented by painting, oiling, greasing or varnishing its surface.

3 0
3 years ago
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