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Vlad [161]
3 years ago
13

Experiment 1 lab report

Chemistry
1 answer:
gogolik [260]3 years ago
4 0
What about it? We need information in order to answer it! :)
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When a solution containing 1.4000 g of Ba(NO3)2 and 2.4000 g of HSO3NH2 is boiled, a precipitate forms. One possible identity fo
Georgia [21]

Answer:

See explanation for detailed solution

Explanation:

The balanced reaction equation is Ba(NO3)2 + 2HSO3NH2 → Ba(SO3NH2)2 + 2HNO3

Number of moles of Ba(NO3)2 = 1.4 g/ 261.337 g/mol = 5.36 × 10^-3 moles

From the reaction equation;

1 mole of Ba(NO3)2 yields 1 mole of Ba(SO3NH2)2

5.36 × 10^-3 moles of Ba(NO3)2 yields 5.36 × 10^-3 moles of Ba(SO3NH2)2

For HSO3NH2

Number of moles = 2.4g/97.10 g/mol =0.0247 moles

2 moles of HSO3NH2 yields 1 mole of Ba(SO3NH2)2

0.0247 moles of HSO3NH2 yields 0.0247 ×1/2 = 0.0137 moles

Hence, Ba(NO3)2 is the limiting reactant

The theoretical yield of Ba(SO3NH2)2 is 5.36 × 10^-3 moles × 329.4986 g/mol = 1.766 g

b)

Number of moles = mass/ molar mass

Molar mass = mass/ number of moles

Molar mass = 1.6925 g/5.36 × 10^-3 moles = 315.76 g

3 0
3 years ago
2Na(s) + 2H,O(l) — 2NaOH(?) + H, (g)<br> 0 what is the missing symbol
Klio2033 [76]

Answer:

the symbol that is missing might be 2.

Explanation:

I am not 100% on this, so correct me if I am wrong.

3 0
2 years ago
If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol
igomit [66]

Answer:

\Delta _RH=4x10^{-3}\frac{kJ}{mol}

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}

Best regards.

5 0
3 years ago
Find the mass of 3.00 mol of acetic acid, C2H4O2.
Alexeev081 [22]
Molar mass 

C₂H₄O₂  = 60.0 g/mol

n = mass / molar mass

3.00 = mass / 60.0

m = 3.00 * 60.0

m = 180 g of <span>C₂H₄O₂ 

hope this helps!</span>
5 0
3 years ago
The sea water has 8.0x10^-1 cg of element strontium. Assuming that all strontium could be recovered, how many grams of strontium
PilotLPTM [1.2K]

984 grams of strontium will be recovered from 9.84x10^8 cubic meter of seawater.

Explanation:

From the question data given is :

volume of strontium in sea water= 9.84x10^8 cubic meter

(1 cubic metre = 1000000 ml)

so 9 .84x10^8 cubic meter

 \frac{9 .84x10^8}{1000000}      = 984 ml.

density of sea water = 1 gram/ml

from the formula mass of strontium can be calculated.

density = \frac{mass}{volume}

mass = density x volume

mass = 1 x 984

         = 984 grams of strontium will be recovered.

98400 centigram of strontium will be recovered.

Strontium is an alkaline earth metal and is highly reactive.

4 0
3 years ago
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