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3 years ago

Use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced in each reaction.

1 answer:

5 0

You didn't include much information as to what you were referring too, however I answered a similar question recently to a friend so I'm hoping this is the same question in complete as to yours, if not I apologize.

a.) 2Na(s) + Cl2(g) -----> 2NaCl(s)
Na --> Na+1 is oxidized
Cl2 --> Cl- is reduced

b.) 2HNO3(aq) + 6HI(aq) ----> 2NO(g) + 3I2(s) + 4H2O(l)
N+5 --> N+2 is reduced
I-1 --> I2 is oxidized

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The reaction below shows how silver chloride can be synthesized.

Vesna [10]

$AgNO_{3} _{(aq)} + NaCl_{(aq)} ----\ \textgreater \ NaNO_{3} _{(aq)} + AgCl _{(aq)}$
mole ration of $AgCl _{(aq)} : AgNO_{3} _{(aq)}$ = 1 : 1

∴ if moles of $AgNO_{3} _{(aq)}$ = 15.0 mol
then " " $AgCl _{(aq)}$ = 15.0 mol

∴ the Option 2 is the answer.

5 0

3 years ago

Read 2 more answers

HELPPPP MEEE PLEASEEE!!!!!!

MrRa [10]

Mass =70 ( Mass of protons=1 ,Mass of neutrons =1, Mass of electron =0.0005(can be ignored))
Therefore, 40 +30=70
Charge= -2 ( it is taking in/attracting electrons to its shell) base on the proton number you are able to identify if it is attracting or releasing an electron, if the electron number is more than proton number then it is attracting therefore resulting in a negative charge vice versa for releasing an electron.

5 0

3 years ago

c. The reaction Br2 (l) --> Br2 (g) has ΔH = 30.91 kJ/mol and ΔS = 93.3 J/mol·K. Use this information to show (within close a

egoroff_w [7]

Answer:

The answer to your question is given below.

Explanation:

From the question given above, the following data were obtained:

Br₂ (l) —> Br₂(g)

Enthalpy change (ΔH) = 30.91 KJ/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

Next, we shall convert 30.91 KJ/mol to J/mol. This can be obtained as follow:

1 KJ/mol = 1000 J/mol

Therefore,

30.91 KJ/mol = 30.91 × 1000

30.91 KJ/mol = 30910 J/mol

Thus, 30.91 KJ/mol is equivalent to 30910 J/mol.

Finally, we shall determine the boiling temperature of bromine. This can be obtained as follow:

Enthalpy change (ΔH) = 30910 J/mol

Entropy change (ΔS) = 93.3 J/mol·K

Boiling temperature (T) =?

ΔS = ΔH / T

93.3 = 30910 / T

Cross multiply

93.3 × T = 30910

Divide both side by 93.3

T = 30910 / 93.3

T = 331.29 K

Thus, the boiling temperature of bromine is 331.29 K

6 0

3 years ago

g a package of aluminum foil contains 50 ft2 of foil, which weighs approximately 8.0 oz. aluminum has a density of 2.70 g/cm3. w

VMariaS [17]

Answer:

1.8×10^-2mm or 0.018mm

Explanation:

Density = mass/Volume

But volume= area×thickness

Area= 50ft2= 46451.52cm2

Mass=8oz= 226.7962g

Density= 2.70g/cm3

Thickness= mass/density ×area

= 226.7962/46451.52×2.70= 1.8×10-3cm= 1.8×10^-2mm

8 0

3 years ago

2 H2 + 2 NO → N2 + 2 H2O the observed rate expression, under some conditions, is: rate = k[NO]2 Which of the following mechanism

guapka [62]

Explanation:

Rate law is defined as the rate of a reaction is directly proportional to the concentration of reactants at constant temperature.

$Rate \propto [\text{concentration of reactant}]^{n}$

= k $[\text{concentration of reactant}]^{n}$

where, k = rate constant

n = order of reaction

For the given reaction, $2H_{2} + 2NO \rightarrow N_{2} + 2H_{2}O$

Hence, its rate will be as follows.

Rate = $k[H_{2}][NO]$

Also, it is known that slowest step in a chemical reaction is the rate determining step.

Hence, for the given rate law correct reaction is as follows.

Step 1 : $H_{2} + NO \rightarrow N + H_{2}O$ (slow)

Balancing this equation it becomes $H_{2} + 2NO \rightarrow N_{2}O + H_{2}O$ (slow)

Step 2: $N_{2}O + H_{2} \rightarrow N_{2} + H_{2}O$ (fast)

4 0

3 years ago