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labwork [276]
2 years ago
10

Question 6 (4 points)

Chemistry
2 answers:
pishuonlain [190]2 years ago
8 0

Answer:

20.9%

Explanation:

I took the test i hope this helps:)

Afina-wow [57]2 years ago
5 0

Answer:

B, "20.9%"

Explanation:

I also took the test.

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Please help me this is my fourth attempt.
Vaselesa [24]

Explanation:

CH4 + 4S ---> CS2 + 2H2S

4) 0.75 mol S × (1 mol CS2/4 mol S) = 0.19 mol CS2

5) 3 mol H2S × (1 mol CH4/2 mol H2S) = 1.5 mol CH4

Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

0.32 mol Al × (26.98 g Al/1 mol Al) = 8.6 g Al

7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

= 0.53 mol Fe2O3

Let's look at how much Fe each reactant will produce:

1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

Note that the given amount of Fe2O3 will give us fewer Fe. Therefore, Fe2O3 is the limiting reactant.

8) Al will produce 1.6 mol Fe × (55.845 g Fe/1 mol Fe)

= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

9) Since Fe2O3 is the limiting reactant, the ideal yield of Fe for the reaction is 61 g. If the actual reaction only gave us 25 g Fe. then the percent yield of Fe is

%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

0.45 mol Fe × (2 mol Al/2 mol Fe) = 0.45 mol Al

0.45 mol Al × (26.98 g Al/1 mol Al) = 12 g

Therefore, the leftover amount of Al is

25 g Al - 12 g Al = 13 g Al

8 0
3 years ago
What change would most likely decrease the rate of a chemical reaction? A) maintaining a temperature close to room temperature B
Nutka1998 [239]

Answer:

D) cooling of the reaction mixture

Explanation:

Increase in temperature speedens the rate of a chemical reaction

5 0
3 years ago
6) Determine la concentración de una solución que ha sido preparada añadiendo 80cc de agua a 320cc de solución alcohólica al 20%
Xelga [282]

Answer:

16% v/v es la nueva concentración de alcohol en la solución

Explanation:

El porcentaje volumen/volumen (% v/v) es definido como 100 veces la relación entre el volumen de soluto (Alcohol en este caso) y el volumen total de la solución (Agua + Alcohol). Para resolver esta pregunta necesitamos hallar el volumen de alcohol y el de agua:

<em>Volumen alcohol:</em>

320cc * (20cc etanol / 100cc) = 64cc etanol

<em>Volumen agua:</em>

80cc + (320cc-64cc) = 336cc agua

<em>% v/v:</em>

64cc / (336cc + 64cc) * 100

= 16% v/v es la nueva concentración de alcohol en la solución

3 0
3 years ago
Identify the oxidizing agent and the reducing agent in the following reactions: (i) 8NH3( g) + 6NO2( g) =&gt; 7N2( g) + 12H2O( l
shusha [124]

Answer:

(i)  Oxidizing Agent: NO2 / Reducing Agent NH3-

(ii) Oxidizing Agent AgNO3 / Reducing Agent Zn

Explanation:

(i) 8NH3( g) + 6NO2( g) => 7N2( g) + 12H2O( l)

In this reaction, both two reactants contain nitrogen with a different oxidation number and produce only one product which contains nitrogen with a unique oxidation state. So, nitrogen is oxidized and reduced in the same reaction.

Nitrogen Undergoes a change in oxidation state from 4+ in NO2 to 0 in N2. It is reduced because it gains electrons (decrease its oxidation state). NO2 is the oxidizing agent (electron acceptor).

Nitrogen Changes from an oxidation state of 3- in NH3 to 0 in N2. It is oxidized because it loses electrons (increase its oxidation state). NH3 is the reducing agent (electron donor)

(ii) Zn(s) +AgNO3(aq) => Zn(NO3)2(aq) + Ag(s)

Ag changes oxidation state from 1+ to 0 in Ag(s).

Ag is reduced because it gains electrons and for this reason and AgNO3 is the oxidizing agent (electron acceptor)

Zn Changes from an oxidation state of 0 in Zn(s) to 2+ in Zn(NO3)2. It is oxidized and for this reason Zn is the reducing agent (electron donor).

Balanced equation:

Zn(s) +2AgNO3(aq) => Zn(NO3)2(aq) + 2Ag(s)

 

4 0
3 years ago
How many grams of sodium are in .500 of a mole
Step2247 [10]

Convert mole to gram by multiplying the molar mass of sodium

0.500mol Na x 22.990g = 11.495g of Na

4 0
3 years ago
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