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fomenos
3 years ago
7

How many grams are in 2.4 moles of NH4?

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
8 0

Answer: 43.2 grams

Explanation:

Given that,

Amount of moles of NH4 (n) = 2.4

Mass of NH4 in grams (m) = ?

For molar mass of NH4, use the molar masses of:

Nitrogen, N = 14g;

Hydrogen, H = 1g

NH4 = 14g + (1g x 4)

= 14g + 4g

= 18g/mol

Since, amount of moles = mass in grams / molar mass

2.4 mole = m / 18g/mol

m = 2.4 mole x 18g/mol

m = 43.2g

Thus, there are 43.2 grams in 2.4 moles of NH4.

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The following data was collected when a reaction was performed experimentally in the laboratory.
REY [17]

Answer:

The answer to your question is 3 moles of AlCl₃

Explanation:

Process

1.- Write and balance the equation

                  Al(NO₃)₃ + 3NaCl   ⇒   3NaNO₃  +  AlCl₃

2.- Determine the limiting reactant

Theoretical proportion     1 mol Al(NO₃)₃ :  3 moles of NaCl                  

Experimental proportion   4 moles Al(NO₃)₃ : 9 moles NaCl

From these values, we determine that the limiting reactant is NaCl because the number of moles increases three times and the number of moles of Al(NO₃)₃  increases four times.

3.- Determine the amount of AlCl₃ using proportions

                       3 moles of NaCl --------------- 1 mol of AlCl₃

                       9 moles of NaCl ----------------  x

                       x = (9 x 1) / 3

                       x = 9 /3

                       x = 3 moles

                     

5 0
3 years ago
3) In the Hydrolysis of Disaccharides and Polysaccharides portion of the lab, starch should give one positive iodine test and on
AfilCa [17]

Answer: potassium iodide is the basic test for starch,and the positive test is blue-black coloration, any other test substance which is not starch will give a negative results.

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Starch is an example of  polysaccharide  and since the standard test for it is potassium iodide solution, it gives a positive test.

Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .

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3 years ago
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The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

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Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

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= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









4 0
3 years ago
Most of the elements of which region of the periodic table are located directly to the right of the metalloids?
scZoUnD [109]

Answer:

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Explanation:

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Andrei [34K]

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6 0
3 years ago
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