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vagabundo [1.1K]
3 years ago
5

Sodium reacts with chlorine gas to form sodium chloride. Write a balanced chemical equation for the reaction, and find the mass

of chlorine gas that will react with 96.6 g of sodium.
Chemistry
2 answers:
weqwewe [10]3 years ago
8 0
2Na + Cl2 ⇒ 2NaCl

96.6g Na ((1 mole Na)/(22.99 g Na))((1 mole Cl2)/(2 mole Na))((70.9 g Cl2)/(1 mole Cl2))=149 g Cl2

Bess [88]3 years ago
4 0
Chemical equation:
2 Na + Cl _{2}   -\ \textgreater \  2NaCl
M ( Na ) = 23 g/moles   ( 2 moles Na = 46 g )
M ( Cl 2 ) = 71 g/moles  ( 1 mole Cl 2 = 71 g )
so Na needs more Cl 2 to react;
96  : x = 46 : 71
46 x = 6816
x = 6816 : 46
x = 148.17 g
Answer: the mass of chlorine gas is 148.17 g.
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A possible mechanism for the gas phase reaction of NO and H2 is as follows: Step 1 2NO N2O2 Step 2 N2O2 + H2 N2O + H2O Step 3 N2
mel-nik [20]

Answer: Option (d) is the correct answer.

Explanation:

Steps involved for the given reaction will be as follows.

Step 1: 2NO \Leftrightarrow N_{2}O_{2}    (fast)

Rate expression for step 1 is as follows.

               Rate = k [NO]^{2}

Step 2: N_{2}O_{2} + H_{2} \rightarrow N_{2}O + H_{2}O

This step 2 is a slow step. Hence, it is a rate determining step.

Step 3. N_{2}O + H_{2} \rightarrow N_{2} + H_{2}O    (fast)

Here, N_{2}O_{2} is intermediate in nature.

All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.

Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.

4 0
3 years ago
Which of the following is a synthetic organic material?
Rudik [331]
The answer is: Plastic
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Hope it helped!
4 0
3 years ago
Read 2 more answers
Does anyone know how to do this ??
lord [1]

Answer:

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Explanation:

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7 0
3 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
3 years ago
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